How do you solve using the quadratic formula: $x^2+x+4=0$ ?

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Answer 1 · 2016-02-16T08:38:32

Solution is $(-1+isqrt15)/2$ and $(-1-isqrt15)/2$

For an equation $ax^2+bx+c=0$ , quadratic formula gives the solution as $(-b+-sqrt(b^2-4ac))/(2a)$ .

As in the given equation $x^2+x+4=0$ , $a=1, b=1$ and $c=4$

Solution is given by $(-1+-sqrt(1^2-4*1.4))/(2*1)$ or

$(-1+-sqrt(-15))/2$ or $(-1+-isqrt15)/2$

Hence solution is $(-1+isqrt15)/2$ and $(-1-isqrt15)/2$

How do you solve using the quadratic formula: x^2+x+4=0x^2+x+4=0?