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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

5x^2-3x-8=0

a = 5; b = -3; c = -8;
Δ = b2-4ac
Δ = -32-4·5·(-8)
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{-(-3)-13}{2*5}=\frac{-10}{10}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+13}{2*5}=\frac{16}{10}
=1+3/5
$