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pav-90 [236]
4 years ago
9

Find the solutions of the equation.

Mathematics
1 answer:
Greeley [361]4 years ago
8 0

Assignment: \bold{Solve \ Equation: \ x^2-4x-5=0}

<><><><><><><>

Answer: \boxed{\bold{x=5,\:x=-1}}

<><><><><><><>

Explanation: \downarrow\downarrow\downarrow

<><><><><><><>

[ Step One ] Solve With Quadric Formula

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

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

<><><><><><><>

\bold{a=1,\:b=-4,\:c=-5:\quad x_{1,\:2}=\frac{-\left(-4\right)\pm \sqrt{\left(-4\right)^2-4\cdot \:1\left(-5\right)}}{2\cdot \:1}}

[ Step Two ] Simplify Expressions

\bold{\frac{-\left(-4\right)+\sqrt{\left(-4\right)^2-4\cdot \:1\cdot \left(-5\right)}}{2\cdot \:1}: \ 5}

\bold{\frac{-\left(-4\right)-\sqrt{\left(-4\right)^2-4\cdot \:1\cdot \left(-5\right)}}{2\cdot \:1}: \ -1}

[ Step Three ] Combine Solutions

\bold{x=5,\:x=-1}

<><><><><><><>

\bold{\rightarrow Mordancy \leftarrow}

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