2x^2+45x-400=0

Solution for 2x^2+45x-400=0 equation:

2x^2+45x-400=0

a = 2; b = 45; c = -400;
Δ = b2-4ac
Δ = 452-4·2·(-400)
Δ = 5225
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:

$\sqrt{\Delta}=\sqrt{5225}=\sqrt{25*209}=\sqrt{25}*\sqrt{209}=5\sqrt{209}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(45)-5\sqrt{209}}{2*2}=\frac{-45-5\sqrt{209}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+5\sqrt{209}}{2*2}=\frac{-45+5\sqrt{209}}{4}
$