How do you solve $H= -0.5x^2 + x + 4$ using the quadratic formula?

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Answer 1 · 2018-04-26T15:36:37

The answer $x_1=1+sqrt3$ and $x_2=1-sqrt3$

show the steps below

$-0.5x^2+x+4=0$

Hit the two side by $-1/0.5$

$x^2-x/0.5-2=0$

$-0.5=-1/2$

$x^2-2x-2=0$

You can solve it by using the law:

$x_1=(-b+sqrt(b^2-4ac))/(2a)$

$x_2=(-b-sqrt(b^2-4ac))/(2a)$

in our formula the values of a b and c equal

$a=1 ,,,,,, b=-2 ,,,,, c=-2$

$x_1=(2+sqrt(4+8))/2$

$x_1=1+sqrt3$

in the same way $x_2$ equal

$x_2=1-sqrt3$

How do you solve H= -0.5x^2 + x + 4H= -0.5x^2 + x + 4 using the quadratic formula?