((4x^2)+62x+58)=((9x^2)-54x+81)

Solution for ((4x^2)+62x+58)=((9x^2)-54x+81) equation:

((4x^2)+62x+58)=((9x^2)-54x+81)

We move all terms to the left:

((4x^2)+62x+58)-(((9x^2)-54x+81))=0

We get rid of parentheses

4x^2+62x-((9x^2-54x+81))+58=0


We calculate terms in parentheses: -((9x^2-54x+81)), so:

(9x^2-54x+81)

We get rid of parentheses

9x^2-54x+81

Back to the equation:

-(9x^2-54x+81)


We get rid of parentheses

4x^2-9x^2+62x+54x-81+58=0

We add all the numbers together, and all the variables

-5x^2+116x-23=0

a = -5; b = 116; c = -23;
Δ = b2-4ac
Δ = 1162-4·(-5)·(-23)
Δ = 12996
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12996}=114$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(116)-114}{2*-5}=\frac{-230}{-10}
=+23
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(116)+114}{2*-5}=\frac{-2}{-10}
=1/5
$