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

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

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

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

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

x = 1, x = 3

Solution steps

- 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

- 3 x > 18

y^{2} - y - 30

- \frac{18}{- 3}

x^{2} + 1 > ∣ - 2 x + 1 ∣

\frac{20 !}{6 ! ( 20 - 6 ) !}

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