r^2+4r-6=0

Solution for r^2+4r-6=0 equation:

r^2+4r-6=0

a = 1; b = 4; c = -6;
Δ = b2-4ac
Δ = 42-4·1·(-6)
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{10}}{2*1}=\frac{-4-2\sqrt{10}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{10}}{2*1}=\frac{-4+2\sqrt{10}}{2}
$