5x2+43x+56=0

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Solution for 5x2+43x+56=0 equation:



5x^2+43x+56=0
a = 5; b = 43; c = +56;
Δ = b2-4ac
Δ = 432-4·5·56
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(43)-27}{2*5}=\frac{-70}{10} =-7 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(43)+27}{2*5}=\frac{-16}{10} =-1+3/5 $

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