17x^2+2x-28=-6+7x^2

Solution for 17x^2+2x-28=-6+7x^2 equation:

17x^2+2x-28=-6+7x^2

We move all terms to the left:

17x^2+2x-28-(-6+7x^2)=0

We get rid of parentheses

17x^2-7x^2+2x+6-28=0

We add all the numbers together, and all the variables

10x^2+2x-22=0

a = 10; b = 2; c = -22;
Δ = b2-4ac
Δ = 22-4·10·(-22)
Δ = 884
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:

$\sqrt{\Delta}=\sqrt{884}=\sqrt{4*221}=\sqrt{4}*\sqrt{221}=2\sqrt{221}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{221}}{2*10}=\frac{-2-2\sqrt{221}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{221}}{2*10}=\frac{-2+2\sqrt{221}}{20}
$