x2+15x-433=0

Solution for x2+15x-433=0 equation:

x2+15x-433=0

We add all the numbers together, and all the variables

x^2+15x-433=0

a = 1; b = 15; c = -433;
Δ = b2-4ac
Δ = 152-4·1·(-433)
Δ = 1957
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-\sqrt{1957}}{2*1}=\frac{-15-\sqrt{1957}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{1957}}{2*1}=\frac{-15+\sqrt{1957}}{2}
$