What are the exact solutions of $x^{2} - 3 x - 7 = 0$ $x^2 − 3x − 7 = 0$ ?

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Answer 1 · 2016-11-13T14:02:56

4.54 and -1.54

$x^{2} - 3 x - 7 = 0$ $x^2-3x-7=0$

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Here

$a = + 1$ $a=+1$

$b = - 3$ $b=-3$

$c = - 7$ $c=-7$

$x = \frac{- (- 3) \pm \sqrt{{(- 3)}^{2} - 4 \times (1) \times (- 7)}}{2 \times (- 1)}$ $x={-(-3)+-sqrt[(-3)^2-4times(1)times(-7)]}/(2times(-1))$

After solving we get

$x = \frac{3 + \sqrt{37}}{2}$ $x={3+sqrt(37)}/(2)$ and $x = \frac{3 - \sqrt{37}}{2}$ $x={3-sqrt(37)}/2$

Therefore $x = 4.54 and x = - 1.54$ $x=4.54 and x=-1.54$

What are the exact solutions of x2−3x−7=0x^2 − 3x − 7 = 0?