Answer:
[tex]\left\{\dfrac{3}{2},\ -\dfrac{5}{3}\right\}[/tex]
Step-by-step explanation:
Given quadratic equation: [tex]6x^2+x-15=0[/tex]
[tex]\implies a=\textsf{6},b=\textsf{1},c=\textsf{-15}[/tex]
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[tex]{\large \textsf{ Quadratic Formula: }}x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\ \Bigg{\|}\ \textsf{when }ax^2+bx+c=0[/tex]
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Step 1: Substitute the given values into the formula and simplify.
[tex]\implies x=\dfrac{-{(\textsf1})\pm \sqrt{\textsf{(1)}^2-4(\textsf{6}){(\textsf{-15})}}}{2{(\textsf{6})}}[/tex]
[tex]\implies x=\dfrac{-1\pm \sqrt{1-4(-90)}}{12}[/tex]
[tex]\implies x=\dfrac{-1\pm \sqrt{1+360}}{12}[/tex]
[tex]\implies x=\dfrac{-1\pm \sqrt{361}}{12}[/tex]
[tex]\implies x=\dfrac{-1\pm 19}{12}[/tex]
Step 2: Separate into two cases.
[tex]x_1=\dfrac{-1+19}{2}[/tex]
[tex]x_2=\dfrac{-1-19}{2}[/tex]
Step 3: Solve both cases.
[tex]\implies x_1=\dfrac{-1+19}{12}=\dfrac{18}{12}=\dfrac{6\times3}{6\times2}=\boxed{\dfrac{3}{2}}[/tex]
[tex]\implies x_2=\dfrac{-1-19}{12}=\dfrac{-20}{12}=\dfrac{-4\times5}{4\times3}=\boxed{-\dfrac{5}{3}}[/tex]
The solutions to this quadratic are: [tex]x_1=\dfrac{3}{2},\ x_2=-\dfrac{5}{3}[/tex]
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