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(x+5)(2x-3)=x^2-13x+15
We move all terms to the left:
(x+5)(2x-3)-(x^2-13x+15)=0
We get rid of parentheses
-x^2+(x+5)(2x-3)+13x-15=0
We multiply parentheses ..
-x^2+(+2x^2-3x+10x-15)+13x-15=0
We add all the numbers together, and all the variables
-1x^2+(+2x^2-3x+10x-15)+13x-15=0
We get rid of parentheses
-1x^2+2x^2-3x+10x+13x-15-15=0
We add all the numbers together, and all the variables
x^2+20x-30=0
a = 1; b = 20; c = -30;
Δ = b2-4ac
Δ = 202-4·1·(-30)
Δ = 520
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{520}=\sqrt{4*130}=\sqrt{4}*\sqrt{130}=2\sqrt{130}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{130}}{2*1}=\frac{-20-2\sqrt{130}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{130}}{2*1}=\frac{-20+2\sqrt{130}}{2} $
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