5x(2+1x)=25(6x+17)

Solution for 5x(2+1x)=25(6x+17) equation:

5x(2+1x)=25(6x+17)

We move all terms to the left:

5x(2+1x)-(25(6x+17))=0

We add all the numbers together, and all the variables

5x(x+2)-(25(6x+17))=0

We multiply parentheses

5x^2+10x-(25(6x+17))=0


We calculate terms in parentheses: -(25(6x+17)), so:

25(6x+17)

We multiply parentheses

150x+425

Back to the equation:

-(150x+425)


We get rid of parentheses

5x^2+10x-150x-425=0

We add all the numbers together, and all the variables

5x^2-140x-425=0

a = 5; b = -140; c = -425;
Δ = b2-4ac
Δ = -1402-4·5·(-425)
Δ = 28100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{28100}=\sqrt{100*281}=\sqrt{100}*\sqrt{281}=10\sqrt{281}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-140)-10\sqrt{281}}{2*5}=\frac{140-10\sqrt{281}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-140)+10\sqrt{281}}{2*5}=\frac{140+10\sqrt{281}}{10}
$