17a-7a(a+1)=4(a-2)

Solution for 17a-7a(a+1)=4(a-2) equation:

17a-7a(a+1)=4(a-2)

We move all terms to the left:

17a-7a(a+1)-(4(a-2))=0

We multiply parentheses

-7a^2+17a-7a-(4(a-2))=0


We calculate terms in parentheses: -(4(a-2)), so:

4(a-2)

We multiply parentheses

4a-8

Back to the equation:

-(4a-8)


We add all the numbers together, and all the variables

-7a^2+10a-(4a-8)=0

We get rid of parentheses

-7a^2+10a-4a+8=0

We add all the numbers together, and all the variables

-7a^2+6a+8=0

a = -7; b = 6; c = +8;
Δ = b2-4ac
Δ = 62-4·(-7)·8
Δ = 260
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{260}=\sqrt{4*65}=\sqrt{4}*\sqrt{65}=2\sqrt{65}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{65}}{2*-7}=\frac{-6-2\sqrt{65}}{-14}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{65}}{2*-7}=\frac{-6+2\sqrt{65}}{-14}
$