4r2+9r=28

Solution for 4r2+9r=28 equation:

4r^2+9r=28

We move all terms to the left:

4r^2+9r-(28)=0

a = 4; b = 9; c = -28;
Δ = b2-4ac
Δ = 92-4·4·(-28)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-23}{2*4}=\frac{-32}{8}
=-4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+23}{2*4}=\frac{14}{8}
=1+3/4
$