4s^2=9s+28

Solution for 4s^2=9s+28 equation:

4s^2=9s+28

We move all terms to the left:

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

We get rid of parentheses

4s^2-9s-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:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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