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The solutions are \( y = -6 \) and \( y = \frac{8}{3} \) (or \( y = 2.6 \)).
Solution
Solve the equation \( 3y^{2}+10y-48=0 \).
Solve the quadratic equation by following steps:
- step0: Solve by factoring:
\(3y^{2}+10y-48=0\)
- step1: Factor the expression:
\(\left(y+6\right)\left(3y-8\right)=0\)
- step2: Separate into possible cases:
\(\begin{align}&3y-8=0\\&y+6=0\end{align}\)
- step3: Solve the equation:
\(\begin{align}&y=\frac{8}{3}\\&y=-6\end{align}\)
- step4: Rewrite:
\(y_{1}=-6,y_{2}=\frac{8}{3}\)
The solutions to the equation \(3y^{2}+10y-48=0\) are \(y=-6\) and \(y=\frac{8}{3}\) or \(y=2.6\).
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Simplify this solution Extra Insights
To solve the quadratic equation \( 3y^{2}+10y-48=0 \), we can use the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 3 \), \( b = 10 \), and \( c = -48 \). Plugging in these values, we first find the discriminant: \[ b^2 - 4ac = 10^2 - 4 \cdot 3 \cdot (-48) = 100 + 576 = 676 \] Then we calculate \( y \): \[ y = \frac{-10 \pm \sqrt{676}}{2 \cdot 3} = \frac{-10 \pm 26}{6} \] This gives us two possible solutions: 1. \( y = \frac{16}{6} = \frac{8}{3} \) 2. \( y = \frac{-36}{6} = -6 \) So, the solutions to the equation \( 3y^{2}+10y-48=0 \) are \( y = \frac{8}{3} \) and \( y = -6 \).
