4(x+3)x=5(x+1)+7

Solution for 4(x+3)x=5(x+1)+7 equation:

4(x+3)x=5(x+1)+7

We move all terms to the left:

4(x+3)x-(5(x+1)+7)=0

We multiply parentheses

4x^2+12x-(5(x+1)+7)=0


We calculate terms in parentheses: -(5(x+1)+7), so:

5(x+1)+7

We multiply parentheses

5x+5+7

We add all the numbers together, and all the variables

5x+12

Back to the equation:

-(5x+12)


We get rid of parentheses

4x^2+12x-5x-12=0

We add all the numbers together, and all the variables

4x^2+7x-12=0

a = 4; b = 7; c = -12;
Δ = b2-4ac
Δ = 72-4·4·(-12)
Δ = 241
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{241}}{2*4}=\frac{-7-\sqrt{241}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{241}}{2*4}=\frac{-7+\sqrt{241}}{8}
$