2x(4x+5)=17

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

2x(4x+5)=17

We move all terms to the left:

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

We multiply parentheses

8x^2+10x-17=0

a = 8; b = 10; c = -17;
Δ = b2-4ac
Δ = 102-4·8·(-17)
Δ = 644
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:

$\sqrt{\Delta}=\sqrt{644}=\sqrt{4*161}=\sqrt{4}*\sqrt{161}=2\sqrt{161}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{161}}{2*8}=\frac{-10-2\sqrt{161}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{161}}{2*8}=\frac{-10+2\sqrt{161}}{16}
$