2r^2-20r+42=0

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Solution for 2r^2-20r+42=0 equation: 



2r^2-20r+42=0

a = 2; b = -20; c = +42;
Δ = b2-4ac
Δ = -202-4·2·42
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$


$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-8}{2*2}=\frac{12}{4}
=3
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+8}{2*2}=\frac{28}{4}
=7
$

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