-q2+3q+5=0

Solution for -q2+3q+5=0 equation:

-q2+3q+5=0

We add all the numbers together, and all the variables

-1q^2+3q+5=0

a = -1; b = 3; c = +5;
Δ = b2-4ac
Δ = 32-4·(-1)·5
Δ = 29
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{-(3)-\sqrt{29}}{2*-1}=\frac{-3-\sqrt{29}}{-2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{29}}{2*-1}=\frac{-3+\sqrt{29}}{-2}
$