2x^2=23-(4x+17)

Solution for 2x^2=23-(4x+17) equation:

2x^2=23-(4x+17)

We move all terms to the left:

2x^2-(23-(4x+17))=0


We calculate terms in parentheses: -(23-(4x+17)), so:

23-(4x+17)

determiningTheFunctionDomain
-(4x+17)+23

We get rid of parentheses

-4x-17+23

We add all the numbers together, and all the variables

-4x+6

Back to the equation:

-(-4x+6)


We get rid of parentheses

2x^2+4x-6=0

a = 2; b = 4; c = -6;
Δ = b2-4ac
Δ = 42-4·2·(-6)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8}{2*2}=\frac{-12}{4}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8}{2*2}=\frac{4}{4}
=1
$