Step-by-step explanation:
Rearrange terms
1
0
=
β
4
+
3
2
10={\color{#c92786}{-4x+3x^{2}}}
10=β4x+3x2
1
0
=
3
2
β
4
10={\color{#c92786}{3x^{2}-4x}}
10=3x2β4x
2
Move terms to the left side
1
0
=
3
2
β
4
10=3x^{2}-4x
10=3x2β4x
1
0
β
(
3
2
β
4
)
=
0
10-\left(3x^{2}-4x\right)=0
10β(3x2β4x)=0
3
Distribute
1
0
β
(
3
2
β
4
)
=
0
10-\left(3x^{2}-4x\right)=0
10β(3x2β4x)=0
1
0
β
3
2
+
4
=
0
10-3x^{2}+4x=0
10β3x2+4x=0
4
Rearrange terms
1
0
β
3
2
+
4
=
0
10-3x^{2}+4x=0
10β3x2+4x=0
β
3
2
+
4
+
1
0
=
0
-3x^{2}+4x+10=0
β3x2+4x+10=0
5
Common factor
β
3
2
+
4
+
1
0
=
0
-3x^{2}+4x+10=0
β3x2+4x+10=0
β
(
3
2
β
4
β
1
0
)
=
0
-\left(3x^{2}-4x-10\right)=0
β(3x2β4xβ10)=0
6
Divide both sides of the equation by the same term
β
(
3
2
β
4
β
1
0
)
=
0
-\left(3x^{2}-4x-10\right)=0
β(3x2β4xβ10)=0
3
2
β
4
β
1
0
=
0
3x^{2}-4x-10=0
3x2β4xβ10=0
7
Use the quadratic formula
=
β
Β±
2
β
4
β
2
x=\frac{-{\color{#e8710a}{b}} \pm \sqrt{{\color{#e8710a}{b}}^{2}-4{\color{#c92786}{a}}{\color{#129eaf}{c{2{\color{#c92786}{a}}}
x=2aβbΒ±b2β4acββ
Once in standard form, identify a, b, and c from the original equation and plug them into the quadratic formula.
3
2
β
4
β
1
0
=
0
3x^{2}-4x-10=0
3x2β4xβ10=0
=
3
a={\color{#c92786}{3}}
a=3
=
β
4
b={\color{#e8710a}{-4}}
b=β4
=
β
1
0
c={\color{#129eaf}{-10}}
c=β10
=
β
(
β
4
)
Β±
(
β
4
)
2
β
4
β
3
(
β
1
0
)
β
2
β
3
x=\frac{-({\color{#e8710a}{-4}}) \pm \sqrt{({\color{#e8710a}{-4}})^{2}-4 \cdot {\color{#c92786}{3}}({\color{#129eaf}{-10}})}}{2 \cdot {\color{#c92786}{3}}}
x=2β
3β(β4)Β±(β4)2β4β
3(β10)ββ
8
Simplify
Evaluate the exponent
Multiply the numbers
Add the numbers
Factorization
Factorization
Factorization
Evaluate the square root
Evaluate the square root
Multiply the numbers
Multiply the numbers
=
4
Β±
2
3
4
β
6
x=\frac{4 \pm 2\sqrt{34}}{6}
x=64Β±234ββ
9
Separate the equations
To solve for the unknown variable, separate into two equations: one with a plus and the other with a minus.
=
4
+
2
3
4
β
6
x=\frac{4+2\sqrt{34}}{6}
x=64+234ββ
=
4
β
2
3
4
β
6
x=\frac{4-2\sqrt{34}}{6}
x=64β234ββ
10
Solve
Rearrange and isolate the variable to find each solution
=
2
+
3
4
β
3
x=\frac{2+\sqrt{34}}{3}
x=32+34ββ
=
2
β
3
4
β
3
x=\frac{2-\sqrt{34}}{3}
x=32β34ββ
=
2
Β±
3
4