Mathematics: What are the zeros of the quadratic function f(x) = 2x2 + 8x – 3? x = –2 – startroot startfraction 11 over 2 endfraction endroot and x = –2 + startroot startfraction 11 over 2 endfraction endroot x = –2 – startroot startfraction 7 over 2 endfraction endroot and x = –2 + startroot startfraction 7 over 2 endfraction endroot x = 2 – startroot startfraction 11 over 2 endfraction endroot and x = 2 + startroot startfraction 11 over 2 endfraction endroot x = 2 – startroot startfraction 7 over 2 endfraction

slope is 2 y-intercept is 1Step-by-step explanation:Im big brain gang...

What are the zeros of the quadratic function f(x)

hanz73

15.01.2022

The zeros are

$x=-2+\frac{\sqrt{22}}{2}$

$x=-2-\frac{\sqrt{22}}{2}$

Step-by-step explanation:

we have

$f(x)=2x^{2}+8x-3$

we know that

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

Equate the function to zero

$2x^{2} +8x-3=0$

$a=2\\b=8\\c=-3$

substitute in the formula

$x=\frac{-8\pm\sqrt{8^{2}-4(2)(-3)}} {2(2)}$

$x=\frac{-8\pm\sqrt{88}} {4}$

$x=\frac{-8\pm2\sqrt{22}} {4}$

$x=-2\pm\frac{\sqrt{22}}{2}$

therefore

$x=-2+\frac{\sqrt{22}}{2}$

$x=-2-\frac{\sqrt{22}}{2}$

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