What are the solutions to the equation 5r^2=-10+27r ?

The solutions to the equation 5r^2=-10+27r are r=5,r= 2/5

Find the zeros of 5r^2=-10+27r

The zeros of 5r^2=-10+27r are r=5,r= 2/5

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5 r^{2} = - 10 + 27 r

r = 5, r = \frac{2}{5}

Decimal Notation

r = 5, r = 0.4

Solution steps

5 r^{2} = - 10 + 27 r

Move

27 r

to the left side

5 r^{2} - 27 r = - 10

Move

10

to the left side

5 r^{2} - 27 r + 10 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 27 ) \pm ( - 27 ) ^{2} - 4 \cdot 5 \cdot 10}{2 \cdot 5}

(- 27)^{2} - 4 \cdot 5 \cdot 10 = 23

r_{1, 2} = \frac{- ( - 27 ) \pm 23}{2 \cdot 5}

Separate the solutions

r_{1} = \frac{- ( - 27 ) + 23}{2 \cdot 5}, r_{2} = \frac{- ( - 27 ) - 23}{2 \cdot 5}

r = \frac{- ( - 27 ) + 23}{2 \cdot 5} : 5

r = \frac{- ( - 27 ) - 23}{2 \cdot 5} : \frac{2}{5}

The solutions to the quadratic equation are:

r = 5, r = \frac{2}{5}

u^{2} + 2 u - 8 = 0

x, x^{2} + x y = 12

4 a^{2} + 9 a = 9

x, f (x + 4) = 3 x^{2} - 6 x

3 x - 9 x^{2} = 0

What are the solutions to the equation 5r^2=-10+27r ?
The solutions to the equation 5r^2=-10+27r are r=5,r= 2/5
Find the zeros of 5r^2=-10+27r
The zeros of 5r^2=-10+27r are r=5,r= 2/5