4x+12=5x^2

Solution for 4x+12=5x^2 equation:

4x+12=5x^2

We move all terms to the left:

4x+12-(5x^2)=0

determiningTheFunctionDomain
-5x^2+4x+12=0

a = -5; b = 4; c = +12;
Δ = b2-4ac
Δ = 42-4·(-5)·12
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-16}{2*-5}=\frac{-20}{-10}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+16}{2*-5}=\frac{12}{-10}
=-1+1/5
$