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

The solutions to the equation 2r^2-3r-4=0 are r=(3+sqrt(41))/4 ,r=(3-sqrt(41))/4

Find the zeros of 2r^2-3r-4=0

The zeros of 2r^2-3r-4=0 are r=(3+sqrt(41))/4 ,r=(3-sqrt(41))/4

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2 r^{2} - 3 r - 4 = 0

r = \frac{3 + 41}{4}, r = \frac{3 - 41}{4}

Decimal Notation

r = 2.35078 \dots, r = - 0.85078 \dots

Solution steps

2 r^{2} - 3 r - 4 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 ( - 4 )}{2 \cdot 2}

(- 3)^{2} - 4 \cdot 2 (- 4) = 41

r_{1, 2} = \frac{- ( - 3 ) \pm 41}{2 \cdot 2}

Separate the solutions

r_{1} = \frac{- ( - 3 ) + 41}{2 \cdot 2}, r_{2} = \frac{- ( - 3 ) - 41}{2 \cdot 2}

r = \frac{- ( - 3 ) + 41}{2 \cdot 2} : \frac{3 + 41}{4}

r = \frac{- ( - 3 ) - 41}{2 \cdot 2} : \frac{3 - 41}{4}

The solutions to the quadratic equation are:

r = \frac{3 + 41}{4}, r = \frac{3 - 41}{4}

2 x^{2} + 8 x - 16 = 0

(2 x + 1) (x - 3) = (x + 1) (x - 2)

(x - 7)^{2} + 2 x = (2 x - 1) (x - 2)

m^{2} - m - 30 = 0

x^{2} - 2 x + 19 = 0

What are the solutions to the equation 2r^2-3r-4=0 ?
The solutions to the equation 2r^2-3r-4=0 are r=(3+sqrt(41))/4 ,r=(3-sqrt(41))/4
Find the zeros of 2r^2-3r-4=0
The zeros of 2r^2-3r-4=0 are r=(3+sqrt(41))/4 ,r=(3-sqrt(41))/4