0.05x^2+15x-6500=0

Solution for 0.05x^2+15x-6500=0 equation:

0.05x^2+15x-6500=0

a = 0.05; b = 15; c = -6500;
Δ = b2-4ac
Δ = 152-4·0.05·(-6500)
Δ = 1525
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:

$\sqrt{\Delta}=\sqrt{1525}=\sqrt{25*61}=\sqrt{25}*\sqrt{61}=5\sqrt{61}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-5\sqrt{61}}{2*0.05}=\frac{-15-5\sqrt{61}}{0.1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+5\sqrt{61}}{2*0.05}=\frac{-15+5\sqrt{61}}{0.1}
$