3x^2+23-14x=4x^2-12+18x

Solution for 3x^2+23-14x=4x^2-12+18x equation:

3x^2+23-14x=4x^2-12+18x

We move all terms to the left:

3x^2+23-14x-(4x^2-12+18x)=0

We get rid of parentheses

3x^2-4x^2-14x-18x+12+23=0

We add all the numbers together, and all the variables

-1x^2-32x+35=0

a = -1; b = -32; c = +35;
Δ = b2-4ac
Δ = -322-4·(-1)·35
Δ = 1164
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1164}=\sqrt{4*291}=\sqrt{4}*\sqrt{291}=2\sqrt{291}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-2\sqrt{291}}{2*-1}=\frac{32-2\sqrt{291}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+2\sqrt{291}}{2*-1}=\frac{32+2\sqrt{291}}{-2}
$