2x2+21x+40=0

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Solution for 2x2+21x+40=0 equation:



2x^2+21x+40=0
a = 2; b = 21; c = +40;
Δ = b2-4ac
Δ = 212-4·2·40
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-11}{2*2}=\frac{-32}{4} =-8 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+11}{2*2}=\frac{-10}{4} =-2+1/2 $

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