(-0.05x^2+17x)=0

Solution for (-0.05x^2+17x)=0 equation:

(-0.05x^2+17x)=0

We get rid of parentheses

-0.05x^2+17x=0

a = -0.05; b = 17; c = 0;
Δ = b2-4ac
Δ = 172-4·(-0.05)·0
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-17}{2*-0.05}=\frac{-34}{-0.1}
=+340
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+17}{2*-0.05}=\frac{0}{-0.1}
=0
$