10(3x+3)=20-30+628x2

Solution for 10(3x+3)=20-30+628x2 equation:

10(3x+3)=20-30+628x^2

We move all terms to the left:

10(3x+3)-(20-30+628x^2)=0

We multiply parentheses

-(20-30+628x^2)+30x+30=0

We get rid of parentheses

-628x^2+30x-20+30+30=0

We add all the numbers together, and all the variables

-628x^2+30x+40=0

a = -628; b = 30; c = +40;
Δ = b2-4ac
Δ = 302-4·(-628)·40
Δ = 101380
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:

$\sqrt{\Delta}=\sqrt{101380}=\sqrt{4*25345}=\sqrt{4}*\sqrt{25345}=2\sqrt{25345}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-2\sqrt{25345}}{2*-628}=\frac{-30-2\sqrt{25345}}{-1256}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+2\sqrt{25345}}{2*-628}=\frac{-30+2\sqrt{25345}}{-1256}
$