(x-3) (2x+5)=0 quadratic equation
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(x+3)(2x-5)=0
2x2+x−15=0
2x2+x-15=0
Use the quadratic formula to find thesolutions.
−b±√b2−4(ac)2a-b±b2-4(ac)2a
Substitute the values a=2a=2, b=1b=1, and c=−15c=-15 into the quadratic formula and solve for xx.
−1±√12−4⋅(2⋅−15)2⋅2-1±12-4⋅(2⋅-15)2⋅2
Simplify the numerator.
x=−1±114x=-1±114
The final answer is the combination of both solutions.
x=52,−3
2x2+x−15=0
2x2+x-15=0
Use the quadratic formula to find thesolutions.
−b±√b2−4(ac)2a-b±b2-4(ac)2a
Substitute the values a=2a=2, b=1b=1, and c=−15c=-15 into the quadratic formula and solve for xx.
−1±√12−4⋅(2⋅−15)2⋅2-1±12-4⋅(2⋅-15)2⋅2
Simplify the numerator.
x=−1±114x=-1±114
The final answer is the combination of both solutions.
x=52,−3
Answered by
4
(x-3)(2x+5)=0
(x-3)=0. or. (2x+5)=0
x=3. or. x=-5/2
(x-3)=0. or. (2x+5)=0
x=3. or. x=-5/2
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