(2x+4)*(5x-1)=17

Solution for (2x+4)*(5x-1)=17 equation:

(2x+4)(5x-1)=17

We move all terms to the left:

(2x+4)(5x-1)-(17)=0

We multiply parentheses ..

(+10x^2-2x+20x-4)-17=0

We get rid of parentheses

10x^2-2x+20x-4-17=0

We add all the numbers together, and all the variables

10x^2+18x-21=0

a = 10; b = 18; c = -21;
Δ = b2-4ac
Δ = 182-4·10·(-21)
Δ = 1164
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:

$\sqrt{\Delta}=\sqrt{1164}=\sqrt{4*291}=\sqrt{4}*\sqrt{291}=2\sqrt{291}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{291}}{2*10}=\frac{-18-2\sqrt{291}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{291}}{2*10}=\frac{-18+2\sqrt{291}}{20}
$