10x^2-50+40=0

Solution for 10x^2-50+40=0 equation:

10x^2-50+40=0

We add all the numbers together, and all the variables

10x^2-10=0

a = 10; b = 0; c = -10;
Δ = b2-4ac
Δ = 02-4·10·(-10)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20}{2*10}=\frac{-20}{20}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20}{2*10}=\frac{20}{20}
=1
$