How do you solve 20x^2 = 13 - x20x2=13−x?

1 Answer
Bounce light knowledge · Shantelle
Apr 7, 2018

x~~0.78,0.83x≈0.78,0.83

Explanation:

Rearranging the equation
=>20x^2+x-13=0 => (1)⇒20x2+x−13=0⇒(1)
ax^2+bx+c=0ax2+bx+c=0=>⇒standard equation

Comparing (11) with standard equation

a=20,b=1,c=-13a=20,b=1,c=−13

Using quadratic formula:

x=(-b+-sqrt(b^2-4ac))/(2a)x=−b±√b2−4ac2a

=>(-1+-sqrt(1^2-4(20×-13)))/(2(20))⇒−1±√12−4(20×−13)2(20)

=>(-1+-sqrt(1-4(-260)))/40⇒−1±√1−4(−260)40

=>(-1+-sqrt(1+1040))/40⇒−1±√1+104040

=>(-1+-sqrt(1041))/40⇒−1±√104140

=>(-1+-32.26)/40⇒−1±32.2640

x~~(-1+32.26)/40,(-1-32.26)/40x≈−1+32.2640,−1−32.2640

x~~31.26/40,-33.26/40x≈31.2640,−33.2640

x~~0.78,0.83x≈0.78,0.83

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