j^2+10j-28=0

Solution for j^2+10j-28=0 equation:

j^2+10j-28=0

a = 1; b = 10; c = -28;
Δ = b2-4ac
Δ = 102-4·1·(-28)
Δ = 212
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{212}=\sqrt{4*53}=\sqrt{4}*\sqrt{53}=2\sqrt{53}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{53}}{2*1}=\frac{-10-2\sqrt{53}}{2}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{53}}{2*1}=\frac{-10+2\sqrt{53}}{2}
$