10x2+20x-800=0

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Solution for 10x2+20x-800=0 equation:



10x^2+20x-800=0
a = 10; b = 20; c = -800;
Δ = b2-4ac
Δ = 202-4·10·(-800)
Δ = 32400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{32400}=180$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-180}{2*10}=\frac{-200}{20} =-10 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+180}{2*10}=\frac{160}{20} =8 $

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