Answer:
[tex]\displaystyle x=\frac{-3+ \sqrt{15}}{2}\approx0.4365\text{ or}\\x=\frac{-3-\sqrt{15}}{2}\approx-3.4365[/tex]
Step-by-step explanation:
We are given the equation:
[tex]-2x^2-6x+3=0[/tex]
And we want to solve it using the quadratic formula:
[tex]\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
First, let's identify our coefficients.
The a is the coefficient in front of the x² term, so a = -2.
b is the coefficient in from of the x term, so b = -6.
And c is the constant so c = 3.
Substitute:
[tex]\displaystyle x=\frac{-(-6)\pm \sqrt{(-6)^2-4(-2)(3)}}{2(-2)}[/tex]
Simplify. -(-6) is just 6. (-6) squared is 36. -4(-2)(3) is 24. And 2(-2) is -4. So:
[tex]\displaystyle x=\frac{6\pm \sqrt{36+24}}{-4}[/tex]
Add:
[tex]\displaystyle x=\frac{6\pm \sqrt{60}}{-4}[/tex]
Simplify the square root:
[tex]\sqrt{60}=\sqrt{4\cdot 15}=\sqrt{4}\cdot\sqrt{15}=2\sqrt{15}[/tex]
So:
[tex]\displaystyle x=\frac{6\pm 2\sqrt{15}}{-4}[/tex]
Simplify by dividing everything by 2:
[tex]\displaystyle x=\frac{3\pm \sqrt{15}}{-2}[/tex]
Move the negative in the denominator to the numerator. The plus/minus sign will not change.
[tex]\displaystyle x=\frac{-3\pm \sqrt{15}}{2}[/tex]
So, our zeros are:
[tex]\displaystyle x=\frac{-3+ \sqrt{15}}{2}\approx0.4365\text{ or}\\x=\frac{-3-\sqrt{15}}{2}\approx-3.4365[/tex]
