Respuesta :

Jisallyouget
2x^2-5x+1=3
2x^2= a -5x=b 1=c
the rule is: ax^2+bx+c= 0
therefore, we have to make the equation equals to zero:
2x^2-5x+1-3=0
2x^2-5x-2= 0
the rule here is: b^2-ac:
delta= 25-16=9
then you square the delta:
=√delta= 3
there are two ways to get x, either:
(-b-delta)÷2a
x...(1)= (5-3)÷4= 1/2
or, (b+delta)÷2a
x...(2)= (5+3)÷4= 2
hope it makes sense.

carlosego

For this case we have the following polynomial:

[tex] 2x^2 - 5x + 1 = 3
[/tex]

Rewriting the polynomial we have:

[tex] 2x^2 - 5x + 1 - 3 = 0

2x^2 - 5x - 2 = 0
[/tex]

Using the quadratic formula to find the solutions we have:

[tex] x = \frac{-b+/-\sqrt{b^2 - 4ac}}{2a} [/tex]

Substituting values:

[tex] x = \frac{-(-5)+/-\sqrt{(-5)^2 - 4(2)(-2)}}{2(2)} [/tex]

Rewriting:

[tex] x = \frac{5+/-\sqrt{25+16}}{4} [/tex]

[tex] x = \frac{5+/-\sqrt{41}}{4} [/tex]

Solution 1:

[tex] x = \frac{5+\sqrt{41}}{4} [/tex]

Solution 2:

[tex] x = \frac{5-\sqrt{41}}{4} [/tex]