Solve the equation. x^{2}+7x+10=0

Weltideepq 2021-11-20 Answered
Solve the equation.
\(\displaystyle{x}^{{{2}}}+{7}{x}+{10}={0}\)

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Expert Answer

Thouturs
Answered 2021-11-21 Author has 7900 answers
Step 1
Given equation is \(\displaystyle{x}^{{{2}}}+{7}{x}+{10}={0}\).
To solve the given equation.
Solution:
Solving the given equation.
\(\displaystyle{x}^{{{2}}}+{7}{x}+{10}={0}\)
\(\displaystyle{x}^{{{2}}}+{5}{x}+{2}{x}+{10}={0}\)
x(x+5)+2(x+5)=0
(x+5)(x+2)=0
(x+5)=0 or (x+2)=0
x=-5 or x=-2
Therefore, solution is x=−5 or x=−2.
Step 2
Hence, required solution is x={-5, -2}.
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Vaing1990
Answered 2021-11-22 Author has 8917 answers
Step 1: Use the formula for the roots of the quadratic equation
\(\displaystyle{x}={\frac{{-{b}\pm\sqrt{{{b}^{{{2}}}-{4}{a}{c}}}}}{{{2}{a}}}}\)
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.
\(\displaystyle{x}^{{{2}}}+{7}{x}+{10}={0}\)
a=1
b=7
c=10
\(\displaystyle{x}={\frac{{-{7}\pm\sqrt{{{7}^{{{2}}}-{4}\cdot{1}\cdot{10}}}}}{{{2}\cdot{1}}}}\)
Step 2: Simplification
\(\displaystyle{x}={\frac{{-{7}\pm\sqrt{{{49}-{4}\cdot{1}\cdot{10}}}}}{{{2}\cdot{1}}}}\)
\(\displaystyle{x}={\frac{{-{7}\pm\sqrt{{{49}-{40}}}}}{{{2}\cdot{1}}}}\)
\(\displaystyle{x}={\frac{{-{7}\pm\sqrt{{{9}}}}}{{{2}\cdot{1}}}}\)
\(\displaystyle{x}={\frac{{-{7}\pm{3}}}{{{2}\cdot{1}}}}\)
\(\displaystyle{x}={\frac{{-{7}\pm{3}}}{{{2}}}}\)
Step 3: Divide the equation
\(\displaystyle{x}={\frac{{-{7}-{3}}}{{{2}}}}\)
Step 4: Calculation
Convert and isolate the variable to find each solution
x=-2
x=-5
The solution
x=-2
x=-5
0

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