21x^2+16x=-3

Solution for 21x^2+16x=-3 equation:

21x^2+16x=-3

We move all terms to the left:

21x^2+16x-(-3)=0

We add all the numbers together, and all the variables

21x^2+16x+3=0

a = 21; b = 16; c = +3;
Δ = b2-4ac
Δ = 162-4·21·3
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2}{2*21}=\frac{-18}{42}
=-3/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2}{2*21}=\frac{-14}{42}
=-1/3
$