(7x^2-8x+10)-(2x^2+x+10)=1

Solution for (7x^2-8x+10)-(2x^2+x+10)=1 equation:

(7x^2-8x+10)-(2x^2+x+10)=1

We move all terms to the left:

(7x^2-8x+10)-(2x^2+x+10)-(1)=0

We get rid of parentheses

7x^2-2x^2-8x-x+10-10-1=0

We add all the numbers together, and all the variables

5x^2-9x-1=0

a = 5; b = -9; c = -1;
Δ = b2-4ac
Δ = -92-4·5·(-1)
Δ = 101
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{-(-9)-\sqrt{101}}{2*5}=\frac{9-\sqrt{101}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{101}}{2*5}=\frac{9+\sqrt{101}}{10}
$