5+(7x(3+7x)-2)=112

Solution for 5+(7x(3+7x)-2)=112 equation:

5+(7x(3+7x)-2)=112

We move all terms to the left:

5+(7x(3+7x)-2)-(112)=0

We add all the numbers together, and all the variables

(7x(7x+3)-2)+5-112=0

We add all the numbers together, and all the variables

(7x(7x+3)-2)-107=0


We calculate terms in parentheses: +(7x(7x+3)-2), so:

7x(7x+3)-2

We multiply parentheses

49x^2+21x-2

Back to the equation:

+(49x^2+21x-2)


We get rid of parentheses

49x^2+21x-2-107=0

We add all the numbers together, and all the variables

49x^2+21x-109=0

a = 49; b = 21; c = -109;
Δ = b2-4ac
Δ = 212-4·49·(-109)
Δ = 21805
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:

$\sqrt{\Delta}=\sqrt{21805}=\sqrt{49*445}=\sqrt{49}*\sqrt{445}=7\sqrt{445}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-7\sqrt{445}}{2*49}=\frac{-21-7\sqrt{445}}{98}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+7\sqrt{445}}{2*49}=\frac{-21+7\sqrt{445}}{98}
$