How do you solve $5 x^{2} - 2 x + 4 = 0$ $5x^2 - 2x +4 = 0$ by quadratic formula?

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How do you solve $5 x^{2} - 2 x + 4 = 0$ $5x^2 - 2x +4 = 0$ by quadratic formula?

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Answer 1 · 2017-07-12T01:00:49

$x = \frac{1 + i \sqrt{19}}{5},$ $x=(1+isqrt19)/5,$ $x = \frac{1 - i \sqrt{19}}{5}$ $x=(1-isqrt19)/5$

Explanation:

Solve by quadratic formula:

$5 x^{2} - 2 x + 4 = 0$ $5x^2-2x+4=0$ is a quadratic equation in standard form: $a x^{2} + b x + c$ $ax^2+bx+c$ , where $a = 5$ $a=5$ , $b = - 2$ $b=-2$ , and $c = 4$ $c=4$ .

quadratic formula:

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

Substitute the values for $a, b, and c$ $a,b, and c$ into the formula.

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

Simplify.

$x = \frac{2 \pm \sqrt{4 - 80}}{10}$ $x=(2+-sqrt(4-80))/10$

$x = \frac{2 \pm \sqrt{- 76}}{10}$ $x=(2+-sqrt(-76))/10$

Prime factorize the number of the square root.

$x = \frac{2 \pm i \sqrt{2 \times 2 \times 19}}{10}$ $x=(2+-isqrt(2xx2xx19))/10$

Simplify.

$x = \frac{2 \pm 2 i \sqrt{19}}{10}$ $x=(2+-2isqrt19)/10$

Reduce.

$x = \frac{1 \pm i \sqrt{19}}{5}$ $x=(1+-isqrt19)/5$

Solutions for $x$ $x$ .

$x = \frac{1 + i \sqrt{19}}{5},$ $x=(1+isqrt19)/5,$ $x = \frac{1 - i \sqrt{19}}{5}$ $x=(1-isqrt19)/5$

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How do you solve $5 x^{2} - 2 x + 4 = 0$ $5x^2 - 2x +4 = 0$ by quadratic formula?

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How do you solve $5 x^{2} - 2 x + 4 = 0$ $5x^2 - 2x +4 = 0$ by quadratic formula?