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\( x^{2}+3 x-40=0 \) |
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Answer» Once in standard form, identify a, b and c from the original equation and plug them into the quadratic formula. \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) x2 + 3x - 40 = 0 a = 1 b = 3 c = -40 \(x = {-3 \pm \sqrt{3^2-4\times1(-40)} \over 2\times1}\) \(x = \frac{-3\pm13}{2}\) \(x = \frac{-3+13}{2}, \frac{-3-13}{2}\) x = 5, -8 |
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