21x^2+69x=28

Solution for 21x^2+69x=28 equation:

21x^2+69x=28

We move all terms to the left:

21x^2+69x-(28)=0

a = 21; b = 69; c = -28;
Δ = b2-4ac
Δ = 692-4·21·(-28)
Δ = 7113
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(69)-\sqrt{7113}}{2*21}=\frac{-69-\sqrt{7113}}{42}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(69)+\sqrt{7113}}{2*21}=\frac{-69+\sqrt{7113}}{42}
$