What are the solutions to the equation 0=-x^2+4x-3 ?

The solutions to the equation 0=-x^2+4x-3 are x=1,x=3

Find the zeros of 0=-x^2+4x-3

The zeros of 0=-x^2+4x-3 are x=1,x=3

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0 = - x^{2} + 4 x - 3

x = 1, x = 3

Solution steps

0 = - x^{2} + 4 x - 3

Switch sides

- x^{2} + 4 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 ( - 1 ) ( - 3 )}{2 ( - 1 )}

4^{2} - 4 (- 1) (- 3) = 2

x_{1, 2} = \frac{- 4 \pm 2}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 4 + 2}{2 ( - 1 )}, x_{2} = \frac{- 4 - 2}{2 ( - 1 )}

x = \frac{- 4 + 2}{2 ( - 1 )} : 1

x = \frac{- 4 - 2}{2 ( - 1 )} : 3

The solutions to the quadratic equation are:

x = 1, x = 3

(x + 2)^{2} = \frac{3}{2}

(3 x + 1)^{2} = x + 2

x, y^{2} = - 6 x^{2} + c

6 x^{2} - 42 x + 24 = 0

x^{2} - 24 x - 81 = 0

What are the solutions to the equation 0=-x^2+4x-3 ?
The solutions to the equation 0=-x^2+4x-3 are x=1,x=3
Find the zeros of 0=-x^2+4x-3
The zeros of 0=-x^2+4x-3 are x=1,x=3