Find the number of $$ x $$-intercepts of the graph of the function. 3. $$ y=-x^{2}+7 x+15 $$ G. One (H.) Two I. None 4. $$ y=3 x^{2}-18 x+27 $$ s. One T. Two U. None Solve the equation using the Quadratic Formula. Round your solution to the nearest tenth, if necessary. 5. $$ x^{2}-9 x+20=0 $$ 6. $$ x^{2}+6 x+5=0 $$ 7. $$ x^{2}-12 x+32=0 $$ 8. $$ 3 x^{2}-5 x+2=0 $$ 9. $$ 1-10 x=-25 x^{2} $$ 10. $$ x^{2}+4 x=7 $$ 11. $$ 8 x^{2}-9=-3 x $$

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3. We find the $$ x $$-intercepts by setting $$ y=0 $$:
$$
-x^2+7x+15=0.
$$
Multiplying both sides by $$-1$$ (which does not change the number of solutions) gives
$$
x^2-7x-15=0.
$$
The discriminant is
$$
\Delta=(-7)^2-4(1)(-15)=49+60=109,
$$
which is positive. Thus, there are two real solutions.  
Answer: Two

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4. Set $$ y=0 $$ to find the $$ x $$-intercepts:
$$
3x^2-18x+27=0.
$$
Divide the entire equation by $$ 3 $$:
$$
x^2-6x+9=0.
$$
Notice that
$$
x^2-6x+9=(x-3)^2=0.
$$
There is a repeated solution, $$ x=3 $$, so the graph has a single $$ x $$-intercept.  
Answer: One

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5. Solve
$$
x^2-9x+20=0.
$$
Here, $$ a=1 $$, $$ b=-9 $$, $$ c=20 $$. The quadratic formula gives
$$
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{9\pm\sqrt{81-80}}{2}=\frac{9\pm1}{2}.
$$
Thus, the solutions are
$$
x=\frac{9+1}{2}=5\quad\text{and}\quad x=\frac{9-1}{2}=4.
$$

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6. Solve
$$
x^2+6x+5=0.
$$
Here, $$ a=1 $$, $$ b=6 $$, $$ c=5 $$.
$$
\Delta=6^2-4(1)(5)=36-20=16.
$$
Applying the quadratic formula:
$$
x=\frac{-6\pm\sqrt{16}}{2}=\frac{-6\pm4}{2}.
$$
The solutions are
$$
x=\frac{-6+4}{2}=-1\quad\text{and}\quad x=\frac{-6-4}{2}=-5.
$$

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7. Solve
$$
x^2-12x+32=0.
$$
Here, $$ a=1 $$, $$ b=-12 $$, $$ c=32 $$ with
$$
\Delta=(-12)^2-4(1)(32)=144-128=16.
$$
Thus,
$$
x=\frac{12\pm\sqrt{16}}{2}=\frac{12\pm4}{2}.
$$
The solutions are
$$
x=\frac{12+4}{2}=8\quad\text{and}\quad x=\frac{12-4}{2}=4.
$$

---

8. Solve
$$
3x^2-5x+2=0.
$$
Here, $$ a=3 $$, $$ b=-5 $$, $$ c=2 $$ and
$$
\Delta=(-5)^2-4(3)(2)=25-24=1.
$$
Then,
$$
x=\frac{5\pm\sqrt{1}}{6}=\frac{5\pm1}{6}.
$$
So,
$$
x=\frac{5+1}{6}=1 \quad\text{and}\quad x=\frac{5-1}{6}=\frac{2}{3}.
$$

---

9. Solve
$$
1-10x=-25x^2.
$$
First, rewrite the equation in standard form:
$$
25x^2-10x+1=0.
$$
Here, $$ a=25 $$, $$ b=-10 $$, $$ c=1 $$ and
$$
\Delta=(-10)^2-4(25)(1)=100-100=0.
$$
A zero discriminant yields one (repeated) solution:
$$
x=\frac{-(-10)}{2\cdot25}=\frac{10}{50}=0.2.
$$

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10. Solve
$$
x^2+4x=7.
$$
Bring all terms to one side:
$$
x^2+4x-7=0.
$$
Here, $$ a=1 $$, $$ b=4 $$, $$ c=-7 $$ and
$$
\Delta=4^2-4(1)(-7)=16+28=44.
$$
Then,
$$
x=\frac{-4\pm\sqrt{44}}{2}=\frac{-4\pm2\sqrt{11}}{2}=-2\pm\sqrt{11}.
$$
Approximating, since $$\sqrt{11}\approx3.3$$:
$$
x\approx -2+3.3=1.3\quad\text{and}\quad x\approx -2-3.3=-5.3.
$$

---

11. Solve
$$
8x^2-9=-3x.
$$
Bring all terms to one side:
$$
8x^2+3x-9=0.
$$
Here, $$ a=8 $$, $$ b=3 $$, $$ c=-9 $$ and
$$
\Delta=3^2-4(8)(-9)=9+288=297.
$$
Thus,
$$
x=\frac{-3\pm\sqrt{297}}{16}.
$$
Recognize that $$\sqrt{297}=3\sqrt{33}$$. Therefore,
$$
x=\frac{-3\pm3\sqrt{33}}{16}.
$$
Approximating, with $$\sqrt{33}\approx5.7$$:
$$
x\approx\frac{-3+17.1}{16}\approx\frac{14.1}{16}\approx0.9,
$$
$$
x\approx\frac{-3-17.1}{16}\approx\frac{-20.1}{16}\approx-1.3.
$$
3. Two 4. One 5. $$ x = 4 $$ and $$ x = 5 $$ 6. $$ x = -5 $$ and $$ x = -1 $$ 7. $$ x = 4 $$ and $$ x = 8 $$ 8. $$ x = \frac{2}{3} $$ and $$ x = 1 $$ 9. $$ x = 0.2 $$ 10. $$ x \approx 1.3 $$ and $$ x \approx -5.3 $$ 11. $$ x \approx 0.9 $$ and $$ x \approx -1.3 $$

Find the number of -intercepts of the graph of the function.
3.
G. One
(H.) Two
I. None
4.
s. One
T. Two
U. None

Solve the equation using the Quadratic Formula. Round your solution to the nearest tenth, if necessary.
5.
6.
7.
8.
9.
10.
11.

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