How do you solve #x² = 2x + 4# using the quadratic formula?

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Answer 1 · 2015-10-17T12:54:52

-1.236, 3.326

Rearrange the formula so all the terms are on one side of the equals sign. (Subtract 2x and 4 from both sides)
#x^2 - 2x - 4 = 0#

The Quadratic Equation is based on the formula form of #y = ax^2 + bx + c#, so now a = 2, b = -2, and c = -4

Substitute these variables into the Quadratic Equation, #x =( (- b - sqrt (b^2 - 4ac)) / (2a))#, #((-b + sqrt(b^2 - 4ac))/(2a))#

#x = (- -2 -sqrt((-2)^2 - 4*1*-4))/(2*1)#

#x = (2-sqrt(4+16))/2 = (2-sqrt(20))/2 = (2-4.472)/2# and

#x = (2+sqrt(4+16))/2 = (2+sqrt(20))/2 = (2+4.472)/2#

#x = -2.472/2 , 6.472/2#