2x^2+3X-62=0

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Solution for 2x^2+3X-62=0 equation: 



2x^2+3x-62=0

a = 2; b = 3; c = -62;
Δ = b2-4ac
Δ = 32-4·2·(-62)
Δ = 505
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{505}}{2*2}=\frac{-3-\sqrt{505}}{4}
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{505}}{2*2}=\frac{-3+\sqrt{505}}{4}
$

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