5(m-3)=4(m-6)m=

Solution for 5(m-3)=4(m-6)m= equation:

5(m-3)=4(m-6)m=

We move all terms to the left:

5(m-3)-(4(m-6)m)=0

We multiply parentheses

5m-(4(m-6)m)-15=0


We calculate terms in parentheses: -(4(m-6)m), so:

4(m-6)m

We multiply parentheses

4m^2-24m

Back to the equation:

-(4m^2-24m)


We get rid of parentheses

-4m^2+5m+24m-15=0

We add all the numbers together, and all the variables

-4m^2+29m-15=0

a = -4; b = 29; c = -15;
Δ = b2-4ac
Δ = 292-4·(-4)·(-15)
Δ = 601
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{-(29)-\sqrt{601}}{2*-4}=\frac{-29-\sqrt{601}}{-8}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+\sqrt{601}}{2*-4}=\frac{-29+\sqrt{601}}{-8}
$