vadulgattp

2021-11-16

Find all real solutions of the equation.

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

George Spencer

Beginner2021-11-17Added 12 answers

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,

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

Step 2

When contrasting the given quadratic equation with its standard form, we obtain

$x=\frac{-70\pm \sqrt{\left({70}^{2}\right)-4\cdot \left(25\right)\cdot \left(49\right)}}{2\cdot 25}$

$=\frac{-70\pm \sqrt{4900-4900}}{50}$

$=\frac{-70}{50}$

$=-\frac{7}{5}$

Therefore, the solution of the given equation is $-\frac{7}{5}$

Witionsion

Beginner2021-11-18Added 19 answers

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

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

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

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

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

$x=-\frac{7}{5}$

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.

a=25

b=70

c=49

Step 2: Raise to the degree

Step 3: Multiply the numbers

Step 4: Subtract the numbers

Step 5: Calculate the square root

Step 6: Add zero

Step 7: Multiply the numbers

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

The solution

$\frac{20b}{{\left(4{b}^{3}\right)}^{3}}$

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