2b^2+6b=10

Solution for 2b^2+6b=10 equation:

2b^2+6b=10

We move all terms to the left:

2b^2+6b-(10)=0

a = 2; b = 6; c = -10;
Δ = b2-4ac
Δ = 62-4·2·(-10)
Δ = 116
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:

$\sqrt{\Delta}=\sqrt{116}=\sqrt{4*29}=\sqrt{4}*\sqrt{29}=2\sqrt{29}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{29}}{2*2}=\frac{-6-2\sqrt{29}}{4}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{29}}{2*2}=\frac{-6+2\sqrt{29}}{4}
$