3 = 7x – 4x^2 If there are no real roots, say so. Huh, I'm confused?

3 = 7x – 4x^2

We rearrange it and get:
4x^2-7x+3=0

Using the formula:
delta=b^2-4ac=(-7)^2-4*4*3=1>0 (that means there are real roots)
root1=(-(-7)+1)/(4*2)=1
root2=(-(-7)-1)/(4*2)=6/8=0.75

This type of equation (ax2 + bx + c = 0) can be solved by the equation

X=(-b +/- sqrt(b2-4ac))/2a

If the term b2-4ac is positive, the equation has real roots. If the term b2-4ac is negative, you have to take the square root of a negative number, and that is not possible with real numbers, which means the equation has no real roots.

It is that simple. That is how you evaluate it. Put the equation in the generic form and evaluate whether b2-4ac is negative (no real roots) or not (real roots).

hey

there are real roots of this equation

x = 3/4 and x = 1

3 = 7x - 4x^2
4x^2 - 7x + 3 = 0
4x^2 - 3x - 4x + 3 = 0
(4x^2 - 3x) - (4x - 3) = 0
x(4x - 3) - 1(4x - 3) = 0
(4x - 3)(x - 1) = 0

4x - 3 = 0
4x = 3
x = 3/4 (0.75)

x - 1 = 0
x = 1

∴ x = 3/4 (0.75) , 1

4x^2-7x+3=0
4x^2-4x-3x+3=0
4x(x-1)-3(x-1)=0
(4x-3)(x-1)=0
x=3/4, x=1.