2x^2-16x+5x-40=-5/2

Solution for 2x^2-16x+5x-40=-5/2 equation:

2x^2-16x+5x-40=-5/2

We move all terms to the left:

2x^2-16x+5x-40-(-5/2)=0

We add all the numbers together, and all the variables

2x^2-11x-40-(-5/2)=0

We get rid of parentheses

2x^2-11x-40+5/2=0

We multiply all the terms by the denominator


2x^2*2-11x*2+5-40*2=0

We add all the numbers together, and all the variables

2x^2*2-11x*2-75=0

Wy multiply elements

4x^2-22x-75=0

a = 4; b = -22; c = -75;
Δ = b2-4ac
Δ = -222-4·4·(-75)
Δ = 1684
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:

$\sqrt{\Delta}=\sqrt{1684}=\sqrt{4*421}=\sqrt{4}*\sqrt{421}=2\sqrt{421}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{421}}{2*4}=\frac{22-2\sqrt{421}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{421}}{2*4}=\frac{22+2\sqrt{421}}{8}
$