7p^2-24p+16=0

Solution for 7p^2-24p+16=0 equation:

7p^2-24p+16=0

a = 7; b = -24; c = +16;
Δ = b2-4ac
Δ = -242-4·7·16
Δ = 128
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-8\sqrt{2}}{2*7}=\frac{24-8\sqrt{2}}{14}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+8\sqrt{2}}{2*7}=\frac{24+8\sqrt{2}}{14}
$