(x2+3)=(3x+7)

Solution for (x2+3)=(3x+7) equation:

(x2+3)=(3x+7)

We move all terms to the left:

(x2+3)-((3x+7))=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

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


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

(3x+7)

We get rid of parentheses

3x+7

Back to the equation:

-(3x+7)


We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2-3x-4=0

a = 1; b = -3; c = -4;
Δ = b2-4ac
Δ = -32-4·1·(-4)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-5}{2*1}=\frac{-2}{2}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+5}{2*1}=\frac{8}{2}
=4
$