1x^2+20x-37=7

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Solution for 1x^2+20x-37=7 equation:



1x^2+20x-37=7
We move all terms to the left:
1x^2+20x-37-(7)=0
We add all the numbers together, and all the variables
x^2+20x-44=0
a = 1; b = 20; c = -44;
Δ = b2-4ac
Δ = 202-4·1·(-44)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-24}{2*1}=\frac{-44}{2} =-22 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+24}{2*1}=\frac{4}{2} =2 $

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