9x^2+24x+16=7(3x+4)

Solution for 9x^2+24x+16=7(3x+4) equation:

9x^2+24x+16=7(3x+4)

We move all terms to the left:

9x^2+24x+16-(7(3x+4))=0


We calculate terms in parentheses: -(7(3x+4)), so:

7(3x+4)

We multiply parentheses

21x+28

Back to the equation:

-(21x+28)


We get rid of parentheses

9x^2+24x-21x-28+16=0

We add all the numbers together, and all the variables

9x^2+3x-12=0

a = 9; b = 3; c = -12;
Δ = b2-4ac
Δ = 32-4·9·(-12)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-21}{2*9}=\frac{-24}{18}
=-1+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+21}{2*9}=\frac{18}{18}
=1
$