r^2-4r-10=0

Solution for r^2-4r-10=0 equation:

r^2-4r-10=0

a = 1; b = -4; c = -10;
Δ = b2-4ac
Δ = -42-4·1·(-10)
Δ = 56
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{14}}{2*1}=\frac{4-2\sqrt{14}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{14}}{2*1}=\frac{4+2\sqrt{14}}{2}
$