Find all real solutions of the equation. 25x^{2}+70x+49=0

Question

Find all real solutions of the equation.

25 x^{2} + 70 x + 49 = 0

Answer 1

Step 1
The given equation is,
$25 x^{2} + 70 x + 49 = 0$
If a quadratic equation is given in the standard form as $a x^{2} + b x + c = 0, a \neq 0,$ then the solutions of the equation is given as,
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
Step 2
On comparing the given quadratic equation with the standard form of the quadratic equation, we get
$x = \frac{- 70 \pm \sqrt{(70^{2}) - 4 \cdot (25) \cdot (49)}}{2 \cdot 25}$
$= \frac{- 70 \pm \sqrt{4900 - 4900}}{50}$
$= \frac{- 70}{50}$
$= - \frac{7}{5}$
Therefore, the solution of the given equation is $- \frac{7}{5}$

Answer 2

Step 1: Use the formula for the roots of the quadratic equation

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

In the standard form, determine a, b and c from the original equation and insert them into the formula for the roots of the quadratic equation.

25 x^{2} + 70 x + 49 = 0

a=25
b=70
c=49

x = \frac{- 70 \pm \sqrt{70^{2} - 4 \cdot 25 \cdot 49}}{2 \cdot 25}

Step 2: Raise to the degree

x = \frac{- 70 \pm \sqrt{4900 - 4 \cdot 25 \cdot 49}}{2 \cdot 25}

Step 3: Multiply the numbers

x = \frac{- 70 \pm \sqrt{4900 - 4900}}{2 \cdot 25}

Step 4: Subtract the numbers

x = \frac{- 70 \pm \sqrt{0}}{2 \cdot 25}

Step 5: Calculate the square root

x = \frac{- 70 \pm 0}{2 \cdot 25}

Step 6: Add zero

x = \frac{- 70}{2 \cdot 25}

Step 7: Multiply the numbers

x = \frac{- 70}{50}

Step 8: Reduce the same terms in the numerator and denominator

x = - \frac{7}{5}

The solution

x = - \frac{7}{5}

Find all real solutions of the equation. 25x^{2}+70x+49=0

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