-2b^2+12b-15=0

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

-2b^2+12b-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:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

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