20x^2+8x-96=0

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Solution for 20x^2+8x-96=0 equation: 



20x^2+8x-96=0

a = 20; b = 8; c = -96;
Δ = b2-4ac
Δ = 82-4·20·(-96)
Δ = 7744
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{7744}=88$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-88}{2*20}=\frac{-96}{40}
=-2+2/5
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+88}{2*20}=\frac{80}{40}
=2
$

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