How do you solve $-7x ^ { 2} + 2x + 4= 0$ ?

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Answer 1 · 2018-01-20T14:45:01

$x~~-0.63,0.91$

The equation is in the form of $ax^2+bx+c=0$ for $a!=0$ , so we can use the quadratic formula, which states that

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

where there are two solutions for $x$ .

Plugging in $a=-7$ , $b=2$ , and $c=4$ , we get

$x=(-2+-sqrt(2^2-(4*-7*4)))/-14$

$x=(-2+-sqrt(4+112))/-14$

$xapprox(-2+-10.77)/-14$

$xapprox 8.77/-14,(-12.77)/-14$

$xapprox -0.63,0.91$

How do you solve -7x ^ { 2} + 2x + 4= 0-7x ^ { 2} + 2x + 4= 0?