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

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Answer 1 · 2018-04-26T15:36:37

The answer $x_{1} = 1 + \sqrt{3}$ $x_1=1+sqrt3$ and $x_{2} = 1 - \sqrt{3}$ $x_2=1-sqrt3$

show the steps below

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

Hit the two side by $- \frac{1}{0.5}$ $-1/0.5$

$x^{2} - \frac{x}{0.5} - 2 = 0$ $x^2-x/0.5-2=0$

$- 0.5 = - \frac{1}{2}$ $-0.5=-1/2$

$x^{2} - 2 x - 2 = 0$ $x^2-2x-2=0$

You can solve it by using the law:

$x_{1} = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a}$ $x_1=(-b+sqrt(b^2-4ac))/(2a)$

$x_{2} = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a}$ $x_2=(-b-sqrt(b^2-4ac))/(2a)$

in our formula the values of a b and c equal

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

$x_{1} = \frac{2 + \sqrt{4 + 8}}{2}$ $x_1=(2+sqrt(4+8))/2$

$x_{1} = 1 + \sqrt{3}$ $x_1=1+sqrt3$

in the same way $x_{2}$ $x_2$ equal

$x_{2} = 1 - \sqrt{3}$ $x_2=1-sqrt3$

How do you solve H=−0.5x2+x+4H= -0.5x^2 + x + 4 using the quadratic formula?