4s(s-3)=5

Solution for 4s(s-3)=5 equation:

4s(s-3)=5

We move all terms to the left:

4s(s-3)-(5)=0

We multiply parentheses

4s^2-12s-5=0

a = 4; b = -12; c = -5;
Δ = b2-4ac
Δ = -122-4·4·(-5)
Δ = 224
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{224}=\sqrt{16*14}=\sqrt{16}*\sqrt{14}=4\sqrt{14}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{14}}{2*4}=\frac{12-4\sqrt{14}}{8}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{14}}{2*4}=\frac{12+4\sqrt{14}}{8}
$