Solving for the values of x, we yield.

[tex]⇢[/tex]For x(+)

[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [/tex] [tex]\begin{gathered}\begin{aligned} x & = \frac{-3 + \sqrt{3^2 - 4(1)(-10)}}{2(1)} \\ & = \frac{-3 + \sqrt{9 + 40}}{2} \\ & = \frac{-3 + \sqrt{49}}{2} \\ & = \frac{-3 + 7}{2} \\ & = \frac{4}{2} \\ & = \boxed{2} \end{aligned}\end{gathered}[/tex]

[tex]\\[/tex]

[tex]⇢[/tex]For x(-)

[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [/tex] [tex]\begin{gathered}\begin{aligned} x & = \frac{-3 - \sqrt{3^2 - 4(1)(-10)}}{2(1)} \\ & = \frac{-3 - \sqrt{9 + 40}}{2} \\ & = \frac{-3 - \sqrt{49}}{2} \\ & = \frac{-3 - 7}{2} \\ & = \frac{-10}{2} \\ & = \boxed{-5} \end{aligned}\end{gathered}[/tex]

Hence, the values of x are 2 and -5.

Verified answer

Answer:

The roots of the equation are 2 and -5.

Step-by-step explanation:

Write the equation ax² + bx + c = 0.

[tex] {x}^{2} + 3x - 10 = 0[/tex]

[tex] \: [/tex]

Identify the values of a, b, c, then plug them into the quadratic formula.

[tex]a = 1[/tex]

[tex]b = 3[/tex]

[tex]c = - 10[/tex]

[tex]x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} \implies x = \frac{ - 3 \pm \sqrt{ {3}^{2} - 4(1)( - 10) } }{2(1)} \\ [/tex]

[tex] \: [/tex]

Simplify the numbers in the quadratic formula.

[tex]x = \frac{ - 3 \pm \sqrt{9 + 40} }{2} \implies x = \frac{ - 3 \pm \sqrt{49} }{2} \\ [/tex]

[tex] \: [/tex]

Simplify the radical and reduce to get the finale answer.

[tex]x = \frac{ - 3 \pm \sqrt{49} }{2} \\ [/tex]

[tex] \: [/tex]

[tex]x = \frac{ - 3 + \sqrt{49} }{2} = \frac{ - 3 + 7}{2} = \frac{4}{2} = 2 \\ [/tex]

or

[tex]x = \frac{ - 3 - \sqrt{49} }{2} = \frac{ - 3 - 7}{2} = \frac{ - 10}{2} = - 5 \\ [/tex]

Therefore, the roots of the equation are 2 and -5.

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