2r^2-12r+15=0

Solution for 2r^2-12r+15=0 equation:

2r^2-12r+15=0

a = 2; b = -12; c = +15;
Δ = b2-4ac
Δ = -122-4·2·15
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{6}}{2*2}=\frac{12-2\sqrt{6}}{4}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{6}}{2*2}=\frac{12+2\sqrt{6}}{4}
$