Solve for z z^2+(k-4)z+0.8=0

Math
z2+(k-4)z+0.8=0
Apply the distributive property.
z2+kz-4z+0.8=0
Use the quadratic formula to find the solutions.
-b±b2-4(ac)2a
Substitute the values a=1, b=k-4, and c=0.8 into the quadratic formula and solve for z.
-(k-4)±(k-4)2-4⋅(1⋅0.8)2⋅1
Simplify.
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Simplify the numerator.
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Apply the distributive property.
z=-k+4±(k-4)2-4⋅(1⋅0.8)2⋅1
Multiply -1 by -4.
z=-k+4±(k-4)2-4⋅(1⋅0.8)2⋅1
Rewrite 4⋅(1⋅0.8) as (2⋅(1⋅0.89442719))2.
z=-k+4±(k-4)2-(2⋅(1⋅0.89442719))22⋅1
Since both terms are perfect squares, factor using the difference of squares formula, a2-b2=(a+b)(a-b) where a=k-4 and b=2⋅(1⋅0.89442719).
z=-k+4±(k-4+2⋅(1⋅0.89442719))(k-4-(2⋅(1⋅0.89442719)))2⋅1
Simplify.
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Multiply 0.89442719 by 1.
z=-k+4±(k-4+2⋅0.89442719)(k-4-(2⋅(1⋅0.89442719)))2⋅1
Multiply 2 by 0.89442719.
z=-k+4±(k-4+1.78885438)(k-4-(2⋅(1⋅0.89442719)))2⋅1
Add -4 and 1.78885438.
z=-k+4±(k-2.21114561)(k-4-(2⋅(1⋅0.89442719)))2⋅1
Multiply 0.89442719 by 1.
z=-k+4±(k-2.21114561)(k-4-(2⋅0.89442719))2⋅1
Multiply 2 by 0.89442719.
z=-k+4±(k-2.21114561)(k-4-1⋅1.78885438)2⋅1
Multiply -1 by 1.78885438.
z=-k+4±(k-2.21114561)(k-4-1.78885438)2⋅1
Subtract 1.78885438 from -4.
z=-k+4±(k-2.21114561)(k-5.78885438)2⋅1
z=-k+4±(k-2.21114561)(k-5.78885438)2⋅1
z=-k+4±(k-2.21114561)(k-5.78885438)2⋅1
Multiply 2 by 1.
z=-k+4±(k-2.21114561)(k-5.78885438)2
z=-k+4±(k-2.21114561)(k-5.78885438)2
The final answer is the combination of both solutions.
z=-k-4-(k-2.21114561)(k-5.78885438)2
z=-k-4+(k-2.21114561)(k-5.78885438)2
Solve for z z^2+(k-4)z+0.8=0

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