20x+5x^2=335

Solution for 20x+5x^2=335 equation:

20x+5x^2=335

We move all terms to the left:

20x+5x^2-(335)=0

a = 5; b = 20; c = -335;
Δ = b2-4ac
Δ = 202-4·5·(-335)
Δ = 7100
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:

$\sqrt{\Delta}=\sqrt{7100}=\sqrt{100*71}=\sqrt{100}*\sqrt{71}=10\sqrt{71}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-10\sqrt{71}}{2*5}=\frac{-20-10\sqrt{71}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+10\sqrt{71}}{2*5}=\frac{-20+10\sqrt{71}}{10}
$