6(7x-2+7)=31x2+20

Solution for 6(7x-2+7)=31x2+20 equation:

6(7x-2+7)=31x^2+20

We move all terms to the left:

6(7x-2+7)-(31x^2+20)=0

We add all the numbers together, and all the variables

6(7x+5)-(31x^2+20)=0

We multiply parentheses

42x-(31x^2+20)+30=0

We get rid of parentheses

-31x^2+42x-20+30=0

We add all the numbers together, and all the variables

-31x^2+42x+10=0

a = -31; b = 42; c = +10;
Δ = b2-4ac
Δ = 422-4·(-31)·10
Δ = 3004
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:

$\sqrt{\Delta}=\sqrt{3004}=\sqrt{4*751}=\sqrt{4}*\sqrt{751}=2\sqrt{751}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{751}}{2*-31}=\frac{-42-2\sqrt{751}}{-62}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{751}}{2*-31}=\frac{-42+2\sqrt{751}}{-62}
$