3s^2+12s-15=0

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Solution for 3s^2+12s-15=0 equation:



3s^2+12s-15=0
a = 3; b = 12; c = -15;
Δ = b2-4ac
Δ = 122-4·3·(-15)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-18}{2*3}=\frac{-30}{6} =-5 $
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+18}{2*3}=\frac{6}{6} =1 $

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