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

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

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

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Witionsion
Step 1: Use the formula for the roots of the quadratic equation
$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
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.
$25{x}^{2}+70x+49=0$
a=25
b=70
c=49
$x=\frac{-70±\sqrt{{70}^{2}-4\cdot 25\cdot 49}}{2\cdot 25}$
Step 2: Raise to the degree
$x=\frac{-70±\sqrt{4900-4\cdot 25\cdot 49}}{2\cdot 25}$
Step 3: Multiply the numbers
$x=\frac{-70±\sqrt{4900-4900}}{2\cdot 25}$
Step 4: Subtract the numbers
$x=\frac{-70±\sqrt{0}}{2\cdot 25}$
Step 5: Calculate the square root
$x=\frac{-70±0}{2\cdot 25}$
$x=\frac{-70}{2\cdot 25}$
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
$x=-\frac{7}{5}$