To solve for the quadratic equation using quadratic formula, the equation must be in general form (ax² + bx + c = 0). Since the given quation is in general form, the values of a, b, and c are
[tex]\begin{aligned} & a = 1 \\ & b = 3 \\ & c = -10 \end{aligned}[/tex]
Substituting the values of a, b, and c, we obtain
[tex]\begin{aligned} x & = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ x & = \frac{-3 \pm \sqrt{3^2 - 4(1)(-10)}}{2(1)} \end{aligned}[/tex]
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SOLUTION:
To solve for the quadratic equation using quadratic formula, the equation must be in general form (ax² + bx + c = 0). Since the given quation is in general form, the values of a, b, and c are
[tex]\begin{aligned} & a = 1 \\ & b = 3 \\ & c = -10 \end{aligned}[/tex]
Substituting the values of a, b, and c, we obtain
[tex]\begin{aligned} x & = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ x & = \frac{-3 \pm \sqrt{3^2 - 4(1)(-10)}}{2(1)} \end{aligned}[/tex]
Solving for the values of x, we yield
• For x(+)
[tex]\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}[/tex]
• For x(-)
[tex]\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}[/tex]
Hence, the values of x are 2 and -5.
[tex]\\[/tex]
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