(m-3)(3m+4)=5m+13

Solution for (m-3)(3m+4)=5m+13 equation:

(m-3)(3m+4)=5m+13

We move all terms to the left:

(m-3)(3m+4)-(5m+13)=0

We get rid of parentheses

(m-3)(3m+4)-5m-13=0

We multiply parentheses ..

(+3m^2+4m-9m-12)-5m-13=0

We get rid of parentheses

3m^2+4m-9m-5m-12-13=0

We add all the numbers together, and all the variables

3m^2-10m-25=0

a = 3; b = -10; c = -25;
Δ = b2-4ac
Δ = -102-4·3·(-25)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-20}{2*3}=\frac{-10}{6}
=-1+2/3
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+20}{2*3}=\frac{30}{6}
=5
$