(4m-7)(m-3)=0

Solution for (4m-7)(m-3)=0 equation:

(4m-7)(m-3)=0

We multiply parentheses ..

(+4m^2-12m-7m+21)=0

We get rid of parentheses

4m^2-12m-7m+21=0

We add all the numbers together, and all the variables

4m^2-19m+21=0

a = 4; b = -19; c = +21;
Δ = b2-4ac
Δ = -192-4·4·21
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-5}{2*4}=\frac{14}{8}
=1+3/4
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+5}{2*4}=\frac{24}{8}
=3
$