2x^2+10x-462=0

Solution for 2x^2+10x-462=0 equation:

2x^2+10x-462=0

a = 2; b = 10; c = -462;
Δ = b2-4ac
Δ = 102-4·2·(-462)
Δ = 3796
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:

$\sqrt{\Delta}=\sqrt{3796}=\sqrt{4*949}=\sqrt{4}*\sqrt{949}=2\sqrt{949}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{949}}{2*2}=\frac{-10-2\sqrt{949}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{949}}{2*2}=\frac{-10+2\sqrt{949}}{4}
$