How do you solve using the quadratic formula $2 x^{2} + 5 x - 3 = 0$ $2x^2 + 5x - 3 = 0$ ?

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How do you solve using the quadratic formula $2 x^{2} + 5 x - 3 = 0$ $2x^2 + 5x - 3 = 0$ ?

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Answer 1 · 2015-05-14T20:57:48

$2 x^{2} + 5 x - 3 = 0$ $2x^2+5x-3 = 0$ is of the form $a x^{2} + b x + c = 0$ $ax^2+bx+c = 0$ with $a = 2$ $a=2$ , $b = 5$ $b=5$ and $c = - 3$ $c=-3$ .

The quadratic formula tells us that this has solutions:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b +- sqrt(b^2-4ac))/(2a)$

$= \frac{- 5 \pm \sqrt{5^{2} - 4 \cdot 2 \cdot (- 3)}}{2 \cdot 2}$ $= (-5 +- sqrt(5^2-4*2*(-3))) / (2*2)$

$= \frac{- 5 \pm \sqrt{25 + 24}}{4}$ $= (-5 +- sqrt(25+24))/4$

$= \frac{- 5 \pm \sqrt{49}}{4}$ $= (-5 +- sqrt(49))/4$

$= \frac{- 5 \pm 7}{4}$ $= (-5 +- 7)/4$

In other words, $x = \frac{1}{2}$ $x = 1/2$ or $x = - 3$ $x = -3$

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