Solve the equation. $x ^ { 2 } + 6 x - 4 = 0$
A) $x = - 3 \pm \sqrt { 13 }$
B) $x = 1 \pm \sqrt { 17 }$
C) $x = - 2 \pm \sqrt { 13 }$
D) $x = - 3 \pm \sqrt { 17 }$
E) The equation has no real roots.

Question

Quizplus · Accepted Answer

This quadratic equation can be solved using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = 6$, and $c = -4$. Plugging these values into the formula gives $x = \frac{-6 \pm \sqrt{(6)^2 - 4(1)(-4)}}{2(1)} = \frac{-6 \pm \sqrt{36 + 16}}{2} = \frac{-6 \pm \sqrt{52}}{2} = -3 \pm \sqrt{13}$.

Solve the Equation x2+6x−4=0x ^ { 2 } + 6 x - 4 = 0x2+6x−4=0

Solve the Equation $x ^ { 2 } + 6 x - 4 = 0$