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What are the roots of the quadratic equation below

Apr 14, 2020

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$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$$

For a quadratic equation : $$ax^2+bx+c=0$$

solution:

$$x = {-15 \pm \sqrt{15^2-4*3*(-5)} \over 2(3)}$$$$= {-15 \pm \sqrt{165} \over 6}$$

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Apr 14, 2020