4x^-5+6x^2=0

Solution for 4x^-5+6x^2=0 equation:

4x^-5+6x^2=0

We add all the numbers together, and all the variables

6x^2+4x-5=0

a = 6; b = 4; c = -5;
Δ = b2-4ac
Δ = 42-4·6·(-5)
Δ = 136
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:

$\sqrt{\Delta}=\sqrt{136}=\sqrt{4*34}=\sqrt{4}*\sqrt{34}=2\sqrt{34}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{34}}{2*6}=\frac{-4-2\sqrt{34}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{34}}{2*6}=\frac{-4+2\sqrt{34}}{12}
$