What are the solutions to the equation R^2+12=7R ?

The solutions to the equation R^2+12=7R are R=4,R=3

Find the zeros of R^2+12=7R

The zeros of R^2+12=7R are R=4,R=3

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R^{2} + 12 = 7 R

R = 4, R = 3

Solution steps

R^{2} + 12 = 7 R

Move

7 R

to the left side

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

R = \frac{- ( - 7 ) + 1}{2 \cdot 1} : 4

R = \frac{- ( - 7 ) - 1}{2 \cdot 1} : 3

The solutions to the quadratic equation are:

R = 4, R = 3

1.7 + k = 5.9

x^{3} - 4 x^{2} + 2 x - 8

\frac{1}{6} + \frac{3}{6}

27 = - 6 n + 9

(x + 4) (x - 3) (1 + x) > 0

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