10x^2+26x=-16

Solution for 10x^2+26x=-16 equation:

10x^2+26x=-16

We move all terms to the left:

10x^2+26x-(-16)=0

We add all the numbers together, and all the variables

10x^2+26x+16=0

a = 10; b = 26; c = +16;
Δ = b2-4ac
Δ = 262-4·10·16
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-6}{2*10}=\frac{-32}{20}
=-1+3/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+6}{2*10}=\frac{-20}{20}
=-1
$