What are the solutions to the equation -x^2+7x-12=0 ?

The solutions to the equation -x^2+7x-12=0 are x=3,x=4

Find the zeros of -x^2+7x-12=0

The zeros of -x^2+7x-12=0 are x=3,x=4

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- x^{2} + 7 x - 12 = 0

x = 3, x = 4

Solution steps

- x^{2} + 7 x - 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 ( - 1 ) ( - 12 )}{2 ( - 1 )}

7^{2} - 4 (- 1) (- 12) = 1

x_{1, 2} = \frac{- 7 \pm 1}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 7 + 1}{2 ( - 1 )}, x_{2} = \frac{- 7 - 1}{2 ( - 1 )}

x = \frac{- 7 + 1}{2 ( - 1 )} : 3

x = \frac{- 7 - 1}{2 ( - 1 )} : 4

The solutions to the quadratic equation are:

x = 3, x = 4

\frac{- 6 \cdot 1 0 ^{- 1}}{- 3 \cdot 1 0 ^{- 2}}

9^{\frac{1}{4}}

\frac{x - 1}{0.2} - \frac{x + 1}{0.5} = 2

x^{2} + 2 x + 11 = 0

- 6 x^{2} + 384 = 0

What are the solutions to the equation -x^2+7x-12=0 ?
The solutions to the equation -x^2+7x-12=0 are x=3,x=4
Find the zeros of -x^2+7x-12=0
The zeros of -x^2+7x-12=0 are x=3,x=4