b(2b-3)=7

Solution for b(2b-3)=7 equation:

b(2b-3)=7

We move all terms to the left:

b(2b-3)-(7)=0

We multiply parentheses

2b^2-3b-7=0

a = 2; b = -3; c = -7;
Δ = b2-4ac
Δ = -32-4·2·(-7)
Δ = 65
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{65}}{2*2}=\frac{3-\sqrt{65}}{4}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{65}}{2*2}=\frac{3+\sqrt{65}}{4}
$