2x2+10x-4=0

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Solution for 2x2+10x-4=0 equation:



2x^2+10x-4=0
a = 2; b = 10; c = -4;
Δ = b2-4ac
Δ = 102-4·2·(-4)
Δ = 132
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{33}}{2*2}=\frac{-10-2\sqrt{33}}{4} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{33}}{2*2}=\frac{-10+2\sqrt{33}}{4} $

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