4x^2=19x-12

Solution for 4x^2=19x-12 equation:

4x^2=19x-12

We move all terms to the left:

4x^2-(19x-12)=0

We get rid of parentheses

4x^2-19x+12=0

a = 4; b = -19; c = +12;
Δ = b2-4ac
Δ = -192-4·4·12
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-13}{2*4}=\frac{6}{8}
=3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+13}{2*4}=\frac{32}{8}
=4
$