5x^2+20x-45=0

Solution for 5x^2+20x-45=0 equation:

5x^2+20x-45=0

a = 5; b = 20; c = -45;
Δ = b2-4ac
Δ = 202-4·5·(-45)
Δ = 1300
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:

$\sqrt{\Delta}=\sqrt{1300}=\sqrt{100*13}=\sqrt{100}*\sqrt{13}=10\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-10\sqrt{13}}{2*5}=\frac{-20-10\sqrt{13}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+10\sqrt{13}}{2*5}=\frac{-20+10\sqrt{13}}{10}
$