# Solve with the quadratic formula. 2p^2-4p=5

$2{p}^{2}-4p=5$
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Raul Garrett
$2{p}^{2}-4p=5$
$2{p}^{2}-4p-5=0$
quadratic formula: $\left(-b±\sqrt{{b}^{2}}-4ac/2a\right)$
b = -4
a = 2
c =-5
$=-\left(-4\right)±{\sqrt{-4}}^{2}-4\left(2\right)\left(-5\right)/2\left(2\right)$
$=4±\sqrt{16}+40/4$
$=\left(4+\sqrt{56}\right)/4$ or $\left(4-\sqrt{56}\right)/4$
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$2{p}^{2}-4p-5=0$
quadratic formula is $x=-b+/-\left({b}^{2}-4ac{\right)}^{.5}/2a$
b=-4
a=2
c=-5
thus
$x=-\left(-4\right)+/-\left(\left(-4{\right)}^{2}-4\ast 2\ast \left(-5\right){\right)}^{.5}/2\ast 2\right)$
$x=\left(4+/-\left(16+40{\right)}^{.5}\right)/4$
$x=\left(4+/-\left(56{\right)}^{.5}\right)/4$
x= 4/4 +/- 7.48/4
x = 1 + 1.87 = 2.87 and x = 1 - 1.87 = -0.87