(7x^2+21x-28)=0

Solution for (7x^2+21x-28)=0 equation:

(7x^2+21x-28)=0

We get rid of parentheses

7x^2+21x-28=0

a = 7; b = 21; c = -28;
Δ = b2-4ac
Δ = 212-4·7·(-28)
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-35}{2*7}=\frac{-56}{14}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+35}{2*7}=\frac{14}{14}
=1
$