6b^2+12b-7=0

Solution for 6b^2+12b-7=0 equation:

6b^2+12b-7=0

a = 6; b = 12; c = -7;
Δ = b2-4ac
Δ = 122-4·6·(-7)
Δ = 312
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:

$\sqrt{\Delta}=\sqrt{312}=\sqrt{4*78}=\sqrt{4}*\sqrt{78}=2\sqrt{78}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{78}}{2*6}=\frac{-12-2\sqrt{78}}{12}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{78}}{2*6}=\frac{-12+2\sqrt{78}}{12}
$