2b^2+10b-28=3

Solution for 2b^2+10b-28=3 equation:

2b^2+10b-28=3

We move all terms to the left:

2b^2+10b-28-(3)=0

We add all the numbers together, and all the variables

2b^2+10b-31=0

a = 2; b = 10; c = -31;
Δ = b2-4ac
Δ = 102-4·2·(-31)
Δ = 348
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:

$\sqrt{\Delta}=\sqrt{348}=\sqrt{4*87}=\sqrt{4}*\sqrt{87}=2\sqrt{87}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{87}}{2*2}=\frac{-10-2\sqrt{87}}{4}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{87}}{2*2}=\frac{-10+2\sqrt{87}}{4}
$