10x^2+40x+400=610

Solution for 10x^2+40x+400=610 equation:

10x^2+40x+400=610

We move all terms to the left:

10x^2+40x+400-(610)=0

We add all the numbers together, and all the variables

10x^2+40x-210=0

a = 10; b = 40; c = -210;
Δ = b2-4ac
Δ = 402-4·10·(-210)
Δ = 10000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{10000}=100$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-100}{2*10}=\frac{-140}{20}
=-7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+100}{2*10}=\frac{60}{20}
=3
$