10x^2+x+-10=-2x+8

Solution for 10x^2+x+-10=-2x+8 equation:

10x^2+x+-10=-2x+8

We move all terms to the left:

10x^2+x+-10-(-2x+8)=0

We add all the numbers together, and all the variables

10x^2+x-(-2x+8)=0

We get rid of parentheses

10x^2+x+2x-8=0

We add all the numbers together, and all the variables

10x^2+3x-8=0

a = 10; b = 3; c = -8;
Δ = b2-4ac
Δ = 32-4·10·(-8)
Δ = 329
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{329}}{2*10}=\frac{-3-\sqrt{329}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{329}}{2*10}=\frac{-3+\sqrt{329}}{20}
$