q^2+10q-37=0

Solution for q^2+10q-37=0 equation:

q^2+10q-37=0

a = 1; b = 10; c = -37;
Δ = b2-4ac
Δ = 102-4·1·(-37)
Δ = 248
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{248}=\sqrt{4*62}=\sqrt{4}*\sqrt{62}=2\sqrt{62}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{62}}{2*1}=\frac{-10-2\sqrt{62}}{2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{62}}{2*1}=\frac{-10+2\sqrt{62}}{2}
$