4m^2=27m-35

Solution for 4m^2=27m-35 equation:

4m^2=27m-35

We move all terms to the left:

4m^2-(27m-35)=0

We get rid of parentheses

4m^2-27m+35=0

a = 4; b = -27; c = +35;
Δ = b2-4ac
Δ = -272-4·4·35
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-13}{2*4}=\frac{14}{8}
=1+3/4
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+13}{2*4}=\frac{40}{8}
=5
$