20x-5x^2=17

Solution for 20x-5x^2=17 equation:

20x-5x^2=17

We move all terms to the left:

20x-5x^2-(17)=0

a = -5; b = 20; c = -17;
Δ = b2-4ac
Δ = 202-4·(-5)·(-17)
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{15}}{2*-5}=\frac{-20-2\sqrt{15}}{-10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{15}}{2*-5}=\frac{-20+2\sqrt{15}}{-10}
$