8n(n+3)=7(n+4)

Solution for 8n(n+3)=7(n+4) equation:

8n(n+3)=7(n+4)

We move all terms to the left:

8n(n+3)-(7(n+4))=0

We multiply parentheses

8n^2+24n-(7(n+4))=0


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

7(n+4)

We multiply parentheses

7n+28

Back to the equation:

-(7n+28)


We get rid of parentheses

8n^2+24n-7n-28=0

We add all the numbers together, and all the variables

8n^2+17n-28=0

a = 8; b = 17; c = -28;
Δ = b2-4ac
Δ = 172-4·8·(-28)
Δ = 1185
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-\sqrt{1185}}{2*8}=\frac{-17-\sqrt{1185}}{16}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+\sqrt{1185}}{2*8}=\frac{-17+\sqrt{1185}}{16}
$