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

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Answer 1 · 2017-07-12T01:00:49

$x=(1+isqrt19)/5,$ $x=(1-isqrt19)/5$

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

quadratic formula:

$x=(-b+-sqrt(b^2-4ac))/(2a)$

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

$x=(-(-2)+-sqrt((-2)^2-4*5*4))/(2*5)$

Simplify.

$x=(2+-sqrt(4-80))/10$

$x=(2+-sqrt(-76))/10$

Prime factorize the number of the square root.

$x=(2+-isqrt(2xx2xx19))/10$

Simplify.

$x=(2+-2isqrt19)/10$

Reduce.

$x=(1+-isqrt19)/5$

Solutions for $x$ .

$x=(1+isqrt19)/5,$ $x=(1-isqrt19)/5$

How do you solve 5x^2 - 2x +4 = 05x^2 - 2x +4 = 0 by quadratic formula?