2x2+12x-62=0

Solution for 2x2+12x-62=0 equation:

2x^2+12x-62=0

a = 2; b = 12; c = -62;
Δ = b2-4ac
Δ = 122-4·2·(-62)
Δ = 640
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:

$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8\sqrt{10}}{2*2}=\frac{-12-8\sqrt{10}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8\sqrt{10}}{2*2}=\frac{-12+8\sqrt{10}}{4}
$