8x^2+69x+40=0

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Solution for 8x^2+69x+40=0 equation:



8x^2+69x+40=0
a = 8; b = 69; c = +40;
Δ = b2-4ac
Δ = 692-4·8·40
Δ = 3481
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3481}=59$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(69)-59}{2*8}=\frac{-128}{16} =-8 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(69)+59}{2*8}=\frac{-10}{16} =-5/8 $

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