(7x+3)=(7x^2+x-18)

Solution for (7x+3)=(7x^2+x-18) equation:

(7x+3)=(7x^2+x-18)

We move all terms to the left:

(7x+3)-((7x^2+x-18))=0

We get rid of parentheses

7x-((7x^2+x-18))+3=0


We calculate terms in parentheses: -((7x^2+x-18)), so:

(7x^2+x-18)

We get rid of parentheses

7x^2+x-18

Back to the equation:

-(7x^2+x-18)


We get rid of parentheses

-7x^2+7x-x+18+3=0

We add all the numbers together, and all the variables

-7x^2+6x+21=0

a = -7; b = 6; c = +21;
Δ = b2-4ac
Δ = 62-4·(-7)·21
Δ = 624
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:

$\sqrt{\Delta}=\sqrt{624}=\sqrt{16*39}=\sqrt{16}*\sqrt{39}=4\sqrt{39}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4\sqrt{39}}{2*-7}=\frac{-6-4\sqrt{39}}{-14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4\sqrt{39}}{2*-7}=\frac{-6+4\sqrt{39}}{-14}
$