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10x^2+35x=(2x+7)
We move all terms to the left:
10x^2+35x-((2x+7))=0
We calculate terms in parentheses: -((2x+7)), so:We get rid of parentheses
(2x+7)
We get rid of parentheses
2x+7
Back to the equation:
-(2x+7)
10x^2+35x-2x-7=0
We add all the numbers together, and all the variables
10x^2+33x-7=0
a = 10; b = 33; c = -7;
Δ = b2-4ac
Δ = 332-4·10·(-7)
Δ = 1369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1369}=37$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-37}{2*10}=\frac{-70}{20} =-3+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+37}{2*10}=\frac{4}{20} =1/5 $
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