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

We add all the numbers together, and all the variables

7x^2-32x+35=0

a = 7; b = -32; c = +35;
Δ = b2-4ac
Δ = -322-4·7·35
Δ = 44
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:

$\sqrt{\Delta}=\sqrt{44}=\sqrt{4*11}=\sqrt{4}*\sqrt{11}=2\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-2\sqrt{11}}{2*7}=\frac{32-2\sqrt{11}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+2\sqrt{11}}{2*7}=\frac{32+2\sqrt{11}}{14}
$