2x^2+16x-62=0

Solution for 2x^2+16x-62=0 equation:

2x^2+16x-62=0

a = 2; b = 16; c = -62;
Δ = b2-4ac
Δ = 162-4·2·(-62)
Δ = 752
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:

$\sqrt{\Delta}=\sqrt{752}=\sqrt{16*47}=\sqrt{16}*\sqrt{47}=4\sqrt{47}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{47}}{2*2}=\frac{-16-4\sqrt{47}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{47}}{2*2}=\frac{-16+4\sqrt{47}}{4}
$