What are the solutions to the equation 7d^2-9d+3=5d^2 ?

The solutions to the equation 7d^2-9d+3=5d^2 are d=(9+sqrt(57))/4 ,d=(9-sqrt(57))/4

Find the zeros of 7d^2-9d+3=5d^2

The zeros of 7d^2-9d+3=5d^2 are d=(9+sqrt(57))/4 ,d=(9-sqrt(57))/4

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7 d^{2} - 9 d + 3 = 5 d^{2}

d = \frac{9 + 57}{4}, d = \frac{9 - 57}{4}

Decimal Notation

d = 4.13745 \dots, d = 0.36254 \dots

Solution steps

7 d^{2} - 9 d + 3 = 5 d^{2}

Move

5 d^{2}

to the left side

2 d^{2} - 9 d + 3 = 0

Solve with the quadratic formula

d_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

(- 9)^{2} - 4 \cdot 2 \cdot 3 = 57

d_{1, 2} = \frac{- ( - 9 ) \pm 57}{2 \cdot 2}

Separate the solutions

d_{1} = \frac{- ( - 9 ) + 57}{2 \cdot 2}, d_{2} = \frac{- ( - 9 ) - 57}{2 \cdot 2}

d = \frac{- ( - 9 ) + 57}{2 \cdot 2} : \frac{9 + 57}{4}

d = \frac{- ( - 9 ) - 57}{2 \cdot 2} : \frac{9 - 57}{4}

The solutions to the quadratic equation are:

d = \frac{9 + 57}{4}, d = \frac{9 - 57}{4}

\frac{w}{2} \leq 2

\frac{- \frac{3}{4}}{\frac{1}{2}}

6 x + 5 \geq 7 x

∣ x - 8 ∣ = 0.02

\frac{u}{5} = 12.5

What are the solutions to the equation 7d^2-9d+3=5d^2 ?
The solutions to the equation 7d^2-9d+3=5d^2 are d=(9+sqrt(57))/4 ,d=(9-sqrt(57))/4
Find the zeros of 7d^2-9d+3=5d^2
The zeros of 7d^2-9d+3=5d^2 are d=(9+sqrt(57))/4 ,d=(9-sqrt(57))/4