Question

What are the solutions of 2x² - 6x+5=0?

Answer 1

No solution

Explanation:

1) Use the quadratic formula

`x=(-b\pm√(b^2-4ac) )/(2a)`

Once in standard form, identify a, b, and c from the original equation and plug them into the quadratic formula.

`2x^2-6x+5=0`

`a=2\\b=-6\\c=5`

`x=(-(-6)\pm√((-6)^2-4*2*5) )/(2*2)`

2) Simplify

Evaluate the exponent:

`x=(6\pm√((-6)^2-4*2*5) )/(2*2)`

`x=(6\pm√(36-4*2*5) )/(2*2)`

Multiply the numbers:

`x=(6\pm√(36-4*2*5) )/(2*2)`

`x=(6\pm√(36-40) )/(2*2)`

Subtract the numbers:

`x=(6\pm√(36-40) )/(2*2)`

`x=(6\pm√(-4) )/(2*2)`

Multiply the numbers

`x=(6\pm√(-4) )/(2*2)`

`x=(6\pm√(-4) )/(4)`

3) No real solutions because the discriminant is negative

The square root of a negative number is not a real number

`d=-4`

Result

No solution

Answer 2

no solutions

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