-q^2+10q-17.5=0

Solution for -q^2+10q-17.5=0 equation:

-q^2+10q-17.5=0

We add all the numbers together, and all the variables

-1q^2+10q-17.5=0

a = -1; b = 10; c = -17.5;
Δ = b2-4ac
Δ = 102-4·(-1)·(-17.5)
Δ = 30
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-\sqrt{30}}{2*-1}=\frac{-10-\sqrt{30}}{-2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+\sqrt{30}}{2*-1}=\frac{-10+\sqrt{30}}{-2}
$