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x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}x=2a−b±b2−4ac
ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0
Completing the square
b2−4acb^2 - 4acb2−4ac
Two solutions (or one repeated solution)
Can be negative
Divide every term by aaa
(b2a)2=b24a2\left(\frac{b}{2a}\right)^2 = \frac{b^2}{4a^2}(2ab)2=4a2b2
x=2x = 2x=2 or x=−4x = -4x=−4
Nature of the solutions
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