4m^2+28m=51

Solution for 4m^2+28m=51 equation:

4m^2+28m=51

We move all terms to the left:

4m^2+28m-(51)=0

a = 4; b = 28; c = -51;
Δ = b2-4ac
Δ = 282-4·4·(-51)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-40}{2*4}=\frac{-68}{8}
=-8+1/2
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+40}{2*4}=\frac{12}{8}
=1+1/2
$