(2x2+4)/(2x+9)=5

Solution for (2x2+4)/(2x+9)=5 equation:

(2x^2+4)/(2x+9)=5

We move all terms to the left:

(2x^2+4)/(2x+9)-(5)=0


Domain of the equation: (2x+9)!=0

We move all terms containing x to the left, all other terms to the right

2x!=-9

x!=-9/2

x!=-4+1/2

x∈R


We multiply all the terms by the denominator


(2x^2+4)-5*(2x+9)=0

We multiply parentheses

(2x^2+4)-10x-45=0

We get rid of parentheses

2x^2-10x+4-45=0

We add all the numbers together, and all the variables

2x^2-10x-41=0

a = 2; b = -10; c = -41;
Δ = b2-4ac
Δ = -102-4·2·(-41)
Δ = 428
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:

$\sqrt{\Delta}=\sqrt{428}=\sqrt{4*107}=\sqrt{4}*\sqrt{107}=2\sqrt{107}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{107}}{2*2}=\frac{10-2\sqrt{107}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{107}}{2*2}=\frac{10+2\sqrt{107}}{4}
$