2x^2-3x-62=0

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}
$