1^3+26x=11x^2+16

Solution for 1^3+26x=11x^2+16 equation:

1^3+26x=11x^2+16

We move all terms to the left:

1^3+26x-(11x^2+16)=0

We add all the numbers together, and all the variables

26x-(11x^2+16)+1=0

We get rid of parentheses

-11x^2+26x-16+1=0

We add all the numbers together, and all the variables

-11x^2+26x-15=0

a = -11; b = 26; c = -15;
Δ = b2-4ac
Δ = 262-4·(-11)·(-15)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-4}{2*-11}=\frac{-30}{-22}
=1+4/11
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+4}{2*-11}=\frac{-22}{-22}
=1
$