8x^2+6x-5=4x-3

Solution for 8x^2+6x-5=4x-3 equation:

8x^2+6x-5=4x-3

We move all terms to the left:

8x^2+6x-5-(4x-3)=0

We get rid of parentheses

8x^2+6x-4x+3-5=0

We add all the numbers together, and all the variables

8x^2+2x-2=0

a = 8; b = 2; c = -2;
Δ = b2-4ac
Δ = 22-4·8·(-2)
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{17}}{2*8}=\frac{-2-2\sqrt{17}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{17}}{2*8}=\frac{-2+2\sqrt{17}}{16}
$