8x+42-2x^2=0

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Solution for 8x+42-2x^2=0 equation: 



8x+42-2x^2=0

a = -2; b = 8; c = +42;
Δ = b2-4ac
Δ = 82-4·(-2)·42
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-20}{2*-2}=\frac{-28}{-4}
=+7
$


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

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