What are the solutions to the equation -3r^2-r-10=-4r^2 ?

The solutions to the equation -3r^2-r-10=-4r^2 are r=(1+sqrt(41))/2 ,r=(1-sqrt(41))/2

Find the zeros of -3r^2-r-10=-4r^2

The zeros of -3r^2-r-10=-4r^2 are r=(1+sqrt(41))/2 ,r=(1-sqrt(41))/2

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- 3 r^{2} - r - 10 = - 4 r^{2}

r = \frac{1 + 41}{2}, r = \frac{1 - 41}{2}

Decimal Notation

r = 3.70156 \dots, r = - 2.70156 \dots

Solution steps

- 3 r^{2} - r - 10 = - 4 r^{2}

Move

4 r^{2}

to the left side

r^{2} - r - 10 = 0

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 1 \cdot (- 10) = 41

r_{1, 2} = \frac{- ( - 1 ) \pm 41}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- ( - 1 ) + 41}{2 \cdot 1}, r_{2} = \frac{- ( - 1 ) - 41}{2 \cdot 1}

r = \frac{- ( - 1 ) + 41}{2 \cdot 1} : \frac{1 + 41}{2}

r = \frac{- ( - 1 ) - 41}{2 \cdot 1} : \frac{1 - 41}{2}

The solutions to the quadratic equation are:

r = \frac{1 + 41}{2}, r = \frac{1 - 41}{2}

x^{2} + 10 x + 25 = 0

2 x^{2} + 4 = - 7 x

8 q^{2} - 26 q + 15 = 0

r^{2} + 1

2 x^{2} - 4 x = 2.5

What are the solutions to the equation -3r^2-r-10=-4r^2 ?
The solutions to the equation -3r^2-r-10=-4r^2 are r=(1+sqrt(41))/2 ,r=(1-sqrt(41))/2
Find the zeros of -3r^2-r-10=-4r^2
The zeros of -3r^2-r-10=-4r^2 are r=(1+sqrt(41))/2 ,r=(1-sqrt(41))/2