7z^2-98z+168=0

Solution for 7z^2-98z+168=0 equation:

7z^2-98z+168=0

a = 7; b = -98; c = +168;
Δ = b2-4ac
Δ = -982-4·7·168
Δ = 4900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4900}=70$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-98)-70}{2*7}=\frac{28}{14}
=2
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-98)+70}{2*7}=\frac{168}{14}
=12
$