12x^2=19x+10

Solution for 12x^2=19x+10 equation:

12x^2=19x+10

We move all terms to the left:

12x^2-(19x+10)=0

We get rid of parentheses

12x^2-19x-10=0

a = 12; b = -19; c = -10;
Δ = b2-4ac
Δ = -192-4·12·(-10)
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-29}{2*12}=\frac{-10}{24}
=-5/12
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+29}{2*12}=\frac{48}{24}
=2
$