2x(2x-3)=10(x+4)

Solution for 2x(2x-3)=10(x+4) equation:

2x(2x-3)=10(x+4)

We move all terms to the left:

2x(2x-3)-(10(x+4))=0

We multiply parentheses

4x^2-6x-(10(x+4))=0


We calculate terms in parentheses: -(10(x+4)), so:

10(x+4)

We multiply parentheses

10x+40

Back to the equation:

-(10x+40)


We get rid of parentheses

4x^2-6x-10x-40=0

We add all the numbers together, and all the variables

4x^2-16x-40=0

a = 4; b = -16; c = -40;
Δ = b2-4ac
Δ = -162-4·4·(-40)
Δ = 896
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:

$\sqrt{\Delta}=\sqrt{896}=\sqrt{64*14}=\sqrt{64}*\sqrt{14}=8\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8\sqrt{14}}{2*4}=\frac{16-8\sqrt{14}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8\sqrt{14}}{2*4}=\frac{16+8\sqrt{14}}{8}
$