Question

Solve the equation z^2+ 3z – 18 = 0.

A. z = –6, z = –3
B. z = –3, z = 6
C. z = 3, z = 6
D. z = –6, z = 3

Answer 1

       ➣ Assignment: $\bold{Solve \ Equation: \ z^2+3z-18=0}$

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       ➣ Answer: $\boxed{\bold{z=3,\:z=-6}}$

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Explanation: ↓↓↓↓↓

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       ➤ [ Step One ] Solve Using Quadratic Formula

$\bold{For\:a\:quadratic\:equation\:of\:the\:form\:ax^2+bx+c=0 \ the\:solutions\:are\:}}$

$\bold{x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}}$

$\bold{a=1,\:b=3,\:c=-18:\quad z_{1,\:2}=\frac{-3\pm \sqrt{3^2-4\cdot \:1\left(-18\right)}}{2\cdot \:1}}$

       ➤ [ Step Two ] Solve Equation

$\bold{\frac{-3+\sqrt{3^2-4\cdot \:1\cdot \left(-18\right)}}{2\cdot \:1}: \ 3}$

       ➤ [ Step Three ] Solve Equation

$\bold{\frac{-3-\sqrt{3^2-4\cdot \:1\cdot \left(-18\right)}}{2\cdot \:1}: \ -6}$

       ➤ [ Step Four ] Combine Solutions

$\bold{z=3,\:z=-6}$

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$\bold{\rightarrow Mordancy \leftarrow}$