7x2+38x+40=0

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Solution for 7x2+38x+40=0 equation:



7x^2+38x+40=0
a = 7; b = 38; c = +40;
Δ = b2-4ac
Δ = 382-4·7·40
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(38)-18}{2*7}=\frac{-56}{14} =-4 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+18}{2*7}=\frac{-20}{14} =-1+3/7 $

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