Answer:
no solutions
Step-by-step explanation:
we use the quadratic formula.
for an equation ax²+bx+c=0, the solutions for x are:
x = (-b±sqrt(b²-4ac))/2a
in our case, a=2, b=-6 and c=5, so:
x= (6±sqrt(36-40))/4
we see that we get a negative number inside the sqrt, so we have no real solutions
Answer:
No solution
Step-by-step explanation:
1) Use the quadratic formula
[tex]x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}[/tex]
Once in standard form, identify a, b, and c from the original equation and plug them into the quadratic formula.
[tex]2x^2-6x+5=0[/tex]
[tex]a=2\\b=-6\\c=5[/tex]
[tex]x=\frac{-(-6)\pm\sqrt{(-6)^2-4*2*5} }{2*2}[/tex]
2) Simplify
Evaluate the exponent:
[tex]x=\frac{6\pm\sqrt{(-6)^2-4*2*5} }{2*2}[/tex]
[tex]x=\frac{6\pm\sqrt{36-4*2*5} }{2*2}[/tex]
Multiply the numbers:
[tex]x=\frac{6\pm\sqrt{36-4*2*5} }{2*2}[/tex]
[tex]x=\frac{6\pm\sqrt{36-40} }{2*2}[/tex]
Subtract the numbers:
[tex]x=\frac{6\pm\sqrt{36-40} }{2*2}[/tex]
[tex]x=\frac{6\pm\sqrt{-4} }{2*2}[/tex]
Multiply the numbers
[tex]x=\frac{6\pm\sqrt{-4} }{2*2}[/tex]
[tex]x=\frac{6\pm\sqrt{-4} }{4}[/tex]
3) No real solutions because the discriminant is negative
The square root of a negative number is not a real number
[tex]d=-4[/tex]
Result
No solution
