Question

Solve the following quadratic equation 2×2-x=21​

Answer 1

$2x^2-x=21\\\\\implies 2x^2 -x -21 =0\\\\\text{Apply quadratic formula},~ x = \dfrac{-b \pm \sqrt{b^2 -4ac}}{2a}\\\\\text{In this case,}~ a =2 , b =-1, c =-21}\\\\\implies x = \dfrac{-(-1) \pm \sqrt{(-1)^2 - 4\cdot 2 \cdot (-21)}}{2(2)}\\\\\implies x =\dfrac{1 \pm \sqrt{169}}{4}\\\\\\\implies x = \dfrac{ 1 \pm 13}4\\\\\\\text{Hence,}\\\\x =\dfrac{1+13}4 = \dfrac{14}4 = \dfrac 72\\\\\text{or}\\\\x =\dfrac{1-13}4 = -\dfrac{12}4 =-3$

Answer 2

Answer:

x1=7/4, x2=-3

Step-by-step explanation:

$2x^{2} - x - 21 = 0\\x1,2 = (1±\sqrt{-168-1})/4\\x1,2 = (1±13)/4\\x1 = 14/4 -> x1 = 7/2\\x2 = -12/4 -> x2 = -3$