(2m+3)(m-1)=4

Solution for (2m+3)(m-1)=4 equation:

(2m+3)(m-1)=4

We move all terms to the left:

(2m+3)(m-1)-(4)=0

We multiply parentheses ..

(+2m^2-2m+3m-3)-4=0

We get rid of parentheses

2m^2-2m+3m-3-4=0

We add all the numbers together, and all the variables

2m^2+m-7=0

a = 2; b = 1; c = -7;
Δ = b2-4ac
Δ = 12-4·2·(-7)
Δ = 57
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{57}}{2*2}=\frac{-1-\sqrt{57}}{4}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{57}}{2*2}=\frac{-1+\sqrt{57}}{4}
$