3x^2+4x-5=7x^2+4x-14

Solution for 3x^2+4x-5=7x^2+4x-14 equation:

3x^2+4x-5=7x^2+4x-14

We move all terms to the left:

3x^2+4x-5-(7x^2+4x-14)=0

We get rid of parentheses

3x^2-7x^2+4x-4x+14-5=0

We add all the numbers together, and all the variables

-4x^2+9=0

a = -4; b = 0; c = +9;
Δ = b2-4ac
Δ = 02-4·(-4)·9
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*-4}=\frac{-12}{-8}
=1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*-4}=\frac{12}{-8}
=-1+1/2
$