(40x^2+67x+28)-(10x^2+9x+2)=0

Solution for (40x^2+67x+28)-(10x^2+9x+2)=0 equation:

(40x^2+67x+28)-(10x^2+9x+2)=0

We get rid of parentheses

40x^2-10x^2+67x-9x+28-2=0

We add all the numbers together, and all the variables

30x^2+58x+26=0

a = 30; b = 58; c = +26;
Δ = b2-4ac
Δ = 582-4·30·26
Δ = 244
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:

$\sqrt{\Delta}=\sqrt{244}=\sqrt{4*61}=\sqrt{4}*\sqrt{61}=2\sqrt{61}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(58)-2\sqrt{61}}{2*30}=\frac{-58-2\sqrt{61}}{60}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(58)+2\sqrt{61}}{2*30}=\frac{-58+2\sqrt{61}}{60}
$