b^2+12b+24=7

Solution for b^2+12b+24=7 equation:

b^2+12b+24=7

We move all terms to the left:

b^2+12b+24-(7)=0

We add all the numbers together, and all the variables

b^2+12b+17=0

a = 1; b = 12; c = +17;
Δ = b2-4ac
Δ = 122-4·1·17
Δ = 76
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:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{19}}{2*1}=\frac{-12-2\sqrt{19}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{19}}{2*1}=\frac{-12+2\sqrt{19}}{2}
$