How do you solve $x^{2} - 5 x - 8 = 0$ $x^2-5x-8=0$ ?

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How do you solve $x^{2} - 5 x - 8 = 0$ $x^2-5x-8=0$ ?

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Answer 1 · 2016-05-10T10:45:49

The roots are $6.275, - 1.275$ $6.275, -1.275$

Explanation:

This is a quadratic equation of form $a x^{2} + b x + c$ $ax^2+bx+c$ where $a = 1; b = - 5; c = - 8; b^{2} - 4 a c = 25 + 32 = 57 (+)$ $a=1;b=-5;c=-8 ; b^2-4ac=25+32=57(+)$ ive. So roots are real. $x = - \frac{b}{2 a} \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a} = 2.5 \pm \frac{\sqrt{57}}{2} = 6.275, - 1.275$ $x= -b/(2a) +- sqrt (b^2-4ac)/(2a)= 2.5 +- sqrt57/2 = 6.275, -1.275$ [Ans]

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How do you solve $x^{2} - 5 x - 8 = 0$ $x^2-5x-8=0$ ?

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How do you solve $x^{2} - 5 x - 8 = 0$ $x^2-5x-8=0$ ?