7x^2-x^2+7x-8=x-3

Solution for 7x^2-x^2+7x-8=x-3 equation:

7x^2-x^2+7x-8=x-3

We move all terms to the left:

7x^2-x^2+7x-8-(x-3)=0

We add all the numbers together, and all the variables

6x^2+7x-(x-3)-8=0

We get rid of parentheses

6x^2+7x-x+3-8=0

We add all the numbers together, and all the variables

6x^2+6x-5=0

a = 6; b = 6; c = -5;
Δ = b2-4ac
Δ = 62-4·6·(-5)
Δ = 156
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:

$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{39}}{2*6}=\frac{-6-2\sqrt{39}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{39}}{2*6}=\frac{-6+2\sqrt{39}}{12}
$