45x^2+435x-6400=0

Solution for 45x^2+435x-6400=0 equation:

45x^2+435x-6400=0

a = 45; b = 435; c = -6400;
Δ = b2-4ac
Δ = 4352-4·45·(-6400)
Δ = 1341225
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:

$\sqrt{\Delta}=\sqrt{1341225}=\sqrt{225*5961}=\sqrt{225}*\sqrt{5961}=15\sqrt{5961}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(435)-15\sqrt{5961}}{2*45}=\frac{-435-15\sqrt{5961}}{90}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(435)+15\sqrt{5961}}{2*45}=\frac{-435+15\sqrt{5961}}{90}
$